For a given quadratic function, identify roots, vertex coordinates, and axis of symmetry [closed]
$begingroup$
For each graph of quadratic functions shown, identify
A) $x^2-3x-4$
B) $2x^2+7x+3$
1) the solutions, or roots
2) coordinates of the vertex
3) equation of axis of symmetry
graphing-functions quadratics conic-sections
edited Nov 24 '18 at 9:34
Blue
48.6k870156
asked Nov 24 '18 at 9:17
Ehsan ZehtabchiEhsan Zehtabchi
263
$endgroup$
closed as off-topic by David G. Stork, Claude Leibovici, Saad, Joel Reyes Noche, Martin R Nov 24 '18 at 11:23
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Claude Leibovici, Saad, Martin R
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
For each graph of quadratic functions shown, identify
A) $x^2-3x-4$
B) $2x^2+7x+3$
1) the solutions, or roots
2) coordinates of the vertex
3) equation of axis of symmetry
graphing-functions quadratics conic-sections
edited Nov 24 '18 at 9:34
Blue
48.6k870156
asked Nov 24 '18 at 9:17
Ehsan ZehtabchiEhsan Zehtabchi
263
$endgroup$
closed as off-topic by David G. Stork, Claude Leibovici, Saad, Joel Reyes Noche, Martin R Nov 24 '18 at 11:23
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Claude Leibovici, Saad, Martin R
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
this is what wolfram alpha gives: wolframalpha.com/input/?i=x%5E2-3x-4 and wolframalpha.com/input/?i=2x%5E2%2B7x%2B3
$endgroup$
– idea
Nov 24 '18 at 9:24
$begingroup$
@idea Not really helpful.
$endgroup$
– Rebellos
Nov 24 '18 at 9:24
$begingroup$
What did you try? Where are you stuck? Do you want roots of each? Do you want intersection points?
$endgroup$
– NoChance
Nov 24 '18 at 9:29
$begingroup$
Welcome (again) to Math.SE! To expand upon a comment your first question received ... The community here prefers/expects questions that include your thoughts, and/or an indication of where you got stuck. Such information helps answerers tailor responses to your skill level, without wasting anyone's time telling you what you already know. (It also helps convince people that you aren't simply trying to get them to do your homework for you.)
$endgroup$
– Blue
Nov 24 '18 at 9:32
$begingroup$
Your question mentions "each graph [...] shown", but shows no graph. You don't (yet) have enough reputation to post an image, but if you can link to an image (preferably hosted on a site such as Imgur.com), someone can add it to your question for you.
$endgroup$
– Blue
Nov 24 '18 at 9:34
add a comment |
$begingroup$
For each graph of quadratic functions shown, identify
A) $x^2-3x-4$
B) $2x^2+7x+3$
1) the solutions, or roots
2) coordinates of the vertex
3) equation of axis of symmetry
graphing-functions quadratics conic-sections
edited Nov 24 '18 at 9:34
Blue
48.6k870156
asked Nov 24 '18 at 9:17
Ehsan ZehtabchiEhsan Zehtabchi
263
$endgroup$
For each graph of quadratic functions shown, identify
A) $x^2-3x-4$
B) $2x^2+7x+3$
1) the solutions, or roots
2) coordinates of the vertex
3) equation of axis of symmetry
graphing-functions quadratics conic-sections
graphing-functions quadratics conic-sections
edited Nov 24 '18 at 9:34
Blue
48.6k870156
asked Nov 24 '18 at 9:17
Ehsan ZehtabchiEhsan Zehtabchi
263
edited Nov 24 '18 at 9:34
Blue
48.6k870156
asked Nov 24 '18 at 9:17
Ehsan ZehtabchiEhsan Zehtabchi
263
edited Nov 24 '18 at 9:34
Blue
48.6k870156
edited Nov 24 '18 at 9:34
Blue
48.6k870156
edited Nov 24 '18 at 9:34
Blue
48.6k870156
48.6k870156
asked Nov 24 '18 at 9:17
Ehsan ZehtabchiEhsan Zehtabchi
263
asked Nov 24 '18 at 9:17
Ehsan ZehtabchiEhsan Zehtabchi
263
asked Nov 24 '18 at 9:17
Ehsan ZehtabchiEhsan Zehtabchi
263
263
closed as off-topic by David G. Stork, Claude Leibovici, Saad, Joel Reyes Noche, Martin R Nov 24 '18 at 11:23
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Claude Leibovici, Saad, Martin R
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by David G. Stork, Claude Leibovici, Saad, Joel Reyes Noche, Martin R Nov 24 '18 at 11:23
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Claude Leibovici, Saad, Martin R
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
this is what wolfram alpha gives: wolframalpha.com/input/?i=x%5E2-3x-4 and wolframalpha.com/input/?i=2x%5E2%2B7x%2B3
$endgroup$
– idea
Nov 24 '18 at 9:24
$begingroup$
@idea Not really helpful.
$endgroup$
– Rebellos
Nov 24 '18 at 9:24
$begingroup$
What did you try? Where are you stuck? Do you want roots of each? Do you want intersection points?
$endgroup$
– NoChance
Nov 24 '18 at 9:29
$begingroup$
Welcome (again) to Math.SE! To expand upon a comment your first question received ... The community here prefers/expects questions that include your thoughts, and/or an indication of where you got stuck. Such information helps answerers tailor responses to your skill level, without wasting anyone's time telling you what you already know. (It also helps convince people that you aren't simply trying to get them to do your homework for you.)
$endgroup$
– Blue
Nov 24 '18 at 9:32
$begingroup$
Your question mentions "each graph [...] shown", but shows no graph. You don't (yet) have enough reputation to post an image, but if you can link to an image (preferably hosted on a site such as Imgur.com), someone can add it to your question for you.
$endgroup$
– Blue
Nov 24 '18 at 9:34
add a comment |
$begingroup$
this is what wolfram alpha gives: wolframalpha.com/input/?i=x%5E2-3x-4 and wolframalpha.com/input/?i=2x%5E2%2B7x%2B3
$endgroup$
– idea
Nov 24 '18 at 9:24
$begingroup$
@idea Not really helpful.
$endgroup$
– Rebellos
Nov 24 '18 at 9:24
$begingroup$
What did you try? Where are you stuck? Do you want roots of each? Do you want intersection points?
$endgroup$
– NoChance
Nov 24 '18 at 9:29
$begingroup$
Welcome (again) to Math.SE! To expand upon a comment your first question received ... The community here prefers/expects questions that include your thoughts, and/or an indication of where you got stuck. Such information helps answerers tailor responses to your skill level, without wasting anyone's time telling you what you already know. (It also helps convince people that you aren't simply trying to get them to do your homework for you.)
$endgroup$
– Blue
Nov 24 '18 at 9:32
$begingroup$
Your question mentions "each graph [...] shown", but shows no graph. You don't (yet) have enough reputation to post an image, but if you can link to an image (preferably hosted on a site such as Imgur.com), someone can add it to your question for you.
$endgroup$
– Blue
Nov 24 '18 at 9:34
$begingroup$
this is what wolfram alpha gives: wolframalpha.com/input/?i=x%5E2-3x-4 and wolframalpha.com/input/?i=2x%5E2%2B7x%2B3
$endgroup$
– idea
Nov 24 '18 at 9:24
$begingroup$
this is what wolfram alpha gives: wolframalpha.com/input/?i=x%5E2-3x-4 and wolframalpha.com/input/?i=2x%5E2%2B7x%2B3
$endgroup$
– idea
Nov 24 '18 at 9:24
$begingroup$
@idea Not really helpful.
$endgroup$
– Rebellos
Nov 24 '18 at 9:24
$begingroup$
@idea Not really helpful.
$endgroup$
– Rebellos
Nov 24 '18 at 9:24
$begingroup$
What did you try? Where are you stuck? Do you want roots of each? Do you want intersection points?
$endgroup$
– NoChance
Nov 24 '18 at 9:29
$begingroup$
What did you try? Where are you stuck? Do you want roots of each? Do you want intersection points?
$endgroup$
– NoChance
Nov 24 '18 at 9:29
$begingroup$
Welcome (again) to Math.SE! To expand upon a comment your first question received ... The community here prefers/expects questions that include your thoughts, and/or an indication of where you got stuck. Such information helps answerers tailor responses to your skill level, without wasting anyone's time telling you what you already know. (It also helps convince people that you aren't simply trying to get them to do your homework for you.)
$endgroup$
– Blue
Nov 24 '18 at 9:32
$begingroup$
Welcome (again) to Math.SE! To expand upon a comment your first question received ... The community here prefers/expects questions that include your thoughts, and/or an indication of where you got stuck. Such information helps answerers tailor responses to your skill level, without wasting anyone's time telling you what you already know. (It also helps convince people that you aren't simply trying to get them to do your homework for you.)
$endgroup$
– Blue
Nov 24 '18 at 9:32
$begingroup$
Your question mentions "each graph [...] shown", but shows no graph. You don't (yet) have enough reputation to post an image, but if you can link to an image (preferably hosted on a site such as Imgur.com), someone can add it to your question for you.
$endgroup$
– Blue
Nov 24 '18 at 9:34
$begingroup$
Your question mentions "each graph [...] shown", but shows no graph. You don't (yet) have enough reputation to post an image, but if you can link to an image (preferably hosted on a site such as Imgur.com), someone can add it to your question for you.
$endgroup$
– Blue
Nov 24 '18 at 9:34
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
The roots of a quadratic equation are found where they give an output of $0$.
Take the first function as an example.
$$f(x) = x^2-3x-4$$
Let $f(x) = 0$.
$$x^2-3x-4 = 0$$
Notice most simple quadratics can be factored. $$x^2-3x-4 = (x-4)(x+1)$$
Now see what values of $x$ allow $f(x) = 0$.
When the function can’t be factored nicely, you can use the Quadratic Formula:
$$x = frac{-bpmsqrt{b^2-4ac}}{2a}$$
To find the coordinates of the vertex $(h, k)$, use
$$h = frac{-b}{2a}$$
$$k = c-frac{b^2}{4a}$$
Most people solve for $h$ and plug the value as $x$ to find $y$ (which becomes $k$), but you can immediately use the formula too.
The axis of symmetry is at $x = h$, which you already find when solving for the vertex.
answered Nov 24 '18 at 9:27
KM101KM101
6,0351525
$endgroup$
$begingroup$
I disagree with giving complete answers to elementary posts that show no effort, whereas hints may motivate the OP into solving the exercise.
$endgroup$
– Rebellos
Nov 24 '18 at 9:29
$begingroup$
I edited the post a bit to just give the basic idea.
$endgroup$
– KM101
Nov 24 '18 at 9:33
$begingroup$
Then +1 ........... !
$endgroup$
– Rebellos
Nov 24 '18 at 9:33
add a comment |
$begingroup$
Hints :
For (1), in order to find the solutions/roots of the expressions given, you need to equate them to $0$ :
$$x^2-3x-4 = 0 $$
$$2x^2+7x-3 = 0$$
To solve the equations above, use the well-known quadratic formula.
For (2), recall that a quadratic polynomial is a parabola sketch graphically, thus its vertex lie on the extrema (maxima or minima) of the parabola. If you're familiar with differentiation, it's enough to find that extreme point by the derivative test. If you're not familiar with differentiation, then there is a stantard formula for the vertices, this is :
$$x_v =-frac{b}{2a}$$
for a quadratic polynomial of the form $ax^2 + bx + c = 0$.
Finally, for (3), since the graphs are parabolas and their vertices are found in (2), what would be the symmetry axis ?
answered Nov 24 '18 at 9:28
RebellosRebellos
14.9k31248
$endgroup$
$begingroup$
"If you're not familiar with differentiation" One can deduce this geometrically. The extremal point has to lie midway between the roots so it's the mean of the roots already found.
$endgroup$
– Winther
Nov 24 '18 at 11:03
$begingroup$
@Winther Well, I mentioned an alternative, which is a standard formula which can also be deduced without the use of differential calculus and is most often that not carried out in pre-calculus elementary algebra courses.
$endgroup$
– Rebellos
Nov 24 '18 at 11:05
add a comment |
$begingroup$
HINT
We have that for $ax^2+bx+c=0$
1) The solutions, or roots
- use that $x_1,x_2=frac{-bpm sqrt{b^2-4ac}}{2a}$
2) Coordinates of the vertex
- use that $x_V=-frac b{2a}$
3) equation of axis of symmetry
- see point $2$
For more details refer to Quadratic equation.
answered Nov 24 '18 at 9:24
gimusigimusi
92.9k84494
$endgroup$
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The roots of a quadratic equation are found where they give an output of $0$.
Take the first function as an example.
$$f(x) = x^2-3x-4$$
Let $f(x) = 0$.
$$x^2-3x-4 = 0$$
Notice most simple quadratics can be factored. $$x^2-3x-4 = (x-4)(x+1)$$
Now see what values of $x$ allow $f(x) = 0$.
When the function can’t be factored nicely, you can use the Quadratic Formula:
$$x = frac{-bpmsqrt{b^2-4ac}}{2a}$$
To find the coordinates of the vertex $(h, k)$, use
$$h = frac{-b}{2a}$$
$$k = c-frac{b^2}{4a}$$
Most people solve for $h$ and plug the value as $x$ to find $y$ (which becomes $k$), but you can immediately use the formula too.
The axis of symmetry is at $x = h$, which you already find when solving for the vertex.
answered Nov 24 '18 at 9:27
KM101KM101
6,0351525
$endgroup$
$begingroup$
I disagree with giving complete answers to elementary posts that show no effort, whereas hints may motivate the OP into solving the exercise.
$endgroup$
– Rebellos
Nov 24 '18 at 9:29
$begingroup$
I edited the post a bit to just give the basic idea.
$endgroup$
– KM101
Nov 24 '18 at 9:33
$begingroup$
Then +1 ........... !
$endgroup$
– Rebellos
Nov 24 '18 at 9:33
add a comment |
$begingroup$
The roots of a quadratic equation are found where they give an output of $0$.
Take the first function as an example.
$$f(x) = x^2-3x-4$$
Let $f(x) = 0$.
$$x^2-3x-4 = 0$$
Notice most simple quadratics can be factored. $$x^2-3x-4 = (x-4)(x+1)$$
Now see what values of $x$ allow $f(x) = 0$.
When the function can’t be factored nicely, you can use the Quadratic Formula:
$$x = frac{-bpmsqrt{b^2-4ac}}{2a}$$
To find the coordinates of the vertex $(h, k)$, use
$$h = frac{-b}{2a}$$
$$k = c-frac{b^2}{4a}$$
Most people solve for $h$ and plug the value as $x$ to find $y$ (which becomes $k$), but you can immediately use the formula too.
The axis of symmetry is at $x = h$, which you already find when solving for the vertex.
answered Nov 24 '18 at 9:27
KM101KM101
6,0351525
$endgroup$
$begingroup$
I disagree with giving complete answers to elementary posts that show no effort, whereas hints may motivate the OP into solving the exercise.
$endgroup$
– Rebellos
Nov 24 '18 at 9:29
$begingroup$
I edited the post a bit to just give the basic idea.
$endgroup$
– KM101
Nov 24 '18 at 9:33
$begingroup$
Then +1 ........... !
$endgroup$
– Rebellos
Nov 24 '18 at 9:33
add a comment |
$begingroup$
The roots of a quadratic equation are found where they give an output of $0$.
Take the first function as an example.
$$f(x) = x^2-3x-4$$
Let $f(x) = 0$.
$$x^2-3x-4 = 0$$
Notice most simple quadratics can be factored. $$x^2-3x-4 = (x-4)(x+1)$$
Now see what values of $x$ allow $f(x) = 0$.
When the function can’t be factored nicely, you can use the Quadratic Formula:
$$x = frac{-bpmsqrt{b^2-4ac}}{2a}$$
To find the coordinates of the vertex $(h, k)$, use
$$h = frac{-b}{2a}$$
$$k = c-frac{b^2}{4a}$$
Most people solve for $h$ and plug the value as $x$ to find $y$ (which becomes $k$), but you can immediately use the formula too.
The axis of symmetry is at $x = h$, which you already find when solving for the vertex.
answered Nov 24 '18 at 9:27
KM101KM101
6,0351525
$endgroup$
The roots of a quadratic equation are found where they give an output of $0$.
Take the first function as an example.
$$f(x) = x^2-3x-4$$
Let $f(x) = 0$.
$$x^2-3x-4 = 0$$
Notice most simple quadratics can be factored. $$x^2-3x-4 = (x-4)(x+1)$$
Now see what values of $x$ allow $f(x) = 0$.
When the function can’t be factored nicely, you can use the Quadratic Formula:
$$x = frac{-bpmsqrt{b^2-4ac}}{2a}$$
To find the coordinates of the vertex $(h, k)$, use
$$h = frac{-b}{2a}$$
$$k = c-frac{b^2}{4a}$$
Most people solve for $h$ and plug the value as $x$ to find $y$ (which becomes $k$), but you can immediately use the formula too.
The axis of symmetry is at $x = h$, which you already find when solving for the vertex.
answered Nov 24 '18 at 9:27
KM101KM101
6,0351525
edited Nov 24 '18 at 9:31
answered Nov 24 '18 at 9:27
KM101KM101
6,0351525
answered Nov 24 '18 at 9:27
KM101KM101
6,0351525
answered Nov 24 '18 at 9:27
KM101KM101
6,0351525
6,0351525
$begingroup$
I disagree with giving complete answers to elementary posts that show no effort, whereas hints may motivate the OP into solving the exercise.
$endgroup$
– Rebellos
Nov 24 '18 at 9:29
$begingroup$
I edited the post a bit to just give the basic idea.
$endgroup$
– KM101
Nov 24 '18 at 9:33
$begingroup$
Then +1 ........... !
$endgroup$
– Rebellos
Nov 24 '18 at 9:33
add a comment |
$begingroup$
I disagree with giving complete answers to elementary posts that show no effort, whereas hints may motivate the OP into solving the exercise.
$endgroup$
– Rebellos
Nov 24 '18 at 9:29
$begingroup$
I edited the post a bit to just give the basic idea.
$endgroup$
– KM101
Nov 24 '18 at 9:33
$begingroup$
Then +1 ........... !
$endgroup$
– Rebellos
Nov 24 '18 at 9:33
$begingroup$
I disagree with giving complete answers to elementary posts that show no effort, whereas hints may motivate the OP into solving the exercise.
$endgroup$
– Rebellos
Nov 24 '18 at 9:29
$begingroup$
I disagree with giving complete answers to elementary posts that show no effort, whereas hints may motivate the OP into solving the exercise.
$endgroup$
– Rebellos
Nov 24 '18 at 9:29
$begingroup$
I edited the post a bit to just give the basic idea.
$endgroup$
– KM101
Nov 24 '18 at 9:33
$begingroup$
I edited the post a bit to just give the basic idea.
$endgroup$
– KM101
Nov 24 '18 at 9:33
$begingroup$
Then +1 ........... !
$endgroup$
– Rebellos
Nov 24 '18 at 9:33
$begingroup$
Then +1 ........... !
$endgroup$
– Rebellos
Nov 24 '18 at 9:33
add a comment |
$begingroup$
Hints :
For (1), in order to find the solutions/roots of the expressions given, you need to equate them to $0$ :
$$x^2-3x-4 = 0 $$
$$2x^2+7x-3 = 0$$
To solve the equations above, use the well-known quadratic formula.
For (2), recall that a quadratic polynomial is a parabola sketch graphically, thus its vertex lie on the extrema (maxima or minima) of the parabola. If you're familiar with differentiation, it's enough to find that extreme point by the derivative test. If you're not familiar with differentiation, then there is a stantard formula for the vertices, this is :
$$x_v =-frac{b}{2a}$$
for a quadratic polynomial of the form $ax^2 + bx + c = 0$.
Finally, for (3), since the graphs are parabolas and their vertices are found in (2), what would be the symmetry axis ?
answered Nov 24 '18 at 9:28
RebellosRebellos
14.9k31248
$endgroup$
$begingroup$
"If you're not familiar with differentiation" One can deduce this geometrically. The extremal point has to lie midway between the roots so it's the mean of the roots already found.
$endgroup$
– Winther
Nov 24 '18 at 11:03
$begingroup$
@Winther Well, I mentioned an alternative, which is a standard formula which can also be deduced without the use of differential calculus and is most often that not carried out in pre-calculus elementary algebra courses.
$endgroup$
– Rebellos
Nov 24 '18 at 11:05
add a comment |
$begingroup$
Hints :
For (1), in order to find the solutions/roots of the expressions given, you need to equate them to $0$ :
$$x^2-3x-4 = 0 $$
$$2x^2+7x-3 = 0$$
To solve the equations above, use the well-known quadratic formula.
For (2), recall that a quadratic polynomial is a parabola sketch graphically, thus its vertex lie on the extrema (maxima or minima) of the parabola. If you're familiar with differentiation, it's enough to find that extreme point by the derivative test. If you're not familiar with differentiation, then there is a stantard formula for the vertices, this is :
$$x_v =-frac{b}{2a}$$
for a quadratic polynomial of the form $ax^2 + bx + c = 0$.
Finally, for (3), since the graphs are parabolas and their vertices are found in (2), what would be the symmetry axis ?
answered Nov 24 '18 at 9:28
RebellosRebellos
14.9k31248
$endgroup$
$begingroup$
"If you're not familiar with differentiation" One can deduce this geometrically. The extremal point has to lie midway between the roots so it's the mean of the roots already found.
$endgroup$
– Winther
Nov 24 '18 at 11:03
$begingroup$
@Winther Well, I mentioned an alternative, which is a standard formula which can also be deduced without the use of differential calculus and is most often that not carried out in pre-calculus elementary algebra courses.
$endgroup$
– Rebellos
Nov 24 '18 at 11:05
add a comment |
$begingroup$
Hints :
For (1), in order to find the solutions/roots of the expressions given, you need to equate them to $0$ :
$$x^2-3x-4 = 0 $$
$$2x^2+7x-3 = 0$$
To solve the equations above, use the well-known quadratic formula.
For (2), recall that a quadratic polynomial is a parabola sketch graphically, thus its vertex lie on the extrema (maxima or minima) of the parabola. If you're familiar with differentiation, it's enough to find that extreme point by the derivative test. If you're not familiar with differentiation, then there is a stantard formula for the vertices, this is :
$$x_v =-frac{b}{2a}$$
for a quadratic polynomial of the form $ax^2 + bx + c = 0$.
Finally, for (3), since the graphs are parabolas and their vertices are found in (2), what would be the symmetry axis ?
answered Nov 24 '18 at 9:28
RebellosRebellos
14.9k31248
$endgroup$
Hints :
For (1), in order to find the solutions/roots of the expressions given, you need to equate them to $0$ :
$$x^2-3x-4 = 0 $$
$$2x^2+7x-3 = 0$$
To solve the equations above, use the well-known quadratic formula.
For (2), recall that a quadratic polynomial is a parabola sketch graphically, thus its vertex lie on the extrema (maxima or minima) of the parabola. If you're familiar with differentiation, it's enough to find that extreme point by the derivative test. If you're not familiar with differentiation, then there is a stantard formula for the vertices, this is :
$$x_v =-frac{b}{2a}$$
for a quadratic polynomial of the form $ax^2 + bx + c = 0$.
Finally, for (3), since the graphs are parabolas and their vertices are found in (2), what would be the symmetry axis ?
answered Nov 24 '18 at 9:28
RebellosRebellos
14.9k31248
edited Nov 24 '18 at 9:35
answered Nov 24 '18 at 9:28
RebellosRebellos
14.9k31248
answered Nov 24 '18 at 9:28
RebellosRebellos
14.9k31248
answered Nov 24 '18 at 9:28
RebellosRebellos
14.9k31248
14.9k31248
$begingroup$
"If you're not familiar with differentiation" One can deduce this geometrically. The extremal point has to lie midway between the roots so it's the mean of the roots already found.
$endgroup$
– Winther
Nov 24 '18 at 11:03
$begingroup$
@Winther Well, I mentioned an alternative, which is a standard formula which can also be deduced without the use of differential calculus and is most often that not carried out in pre-calculus elementary algebra courses.
$endgroup$
– Rebellos
Nov 24 '18 at 11:05
add a comment |
$begingroup$
"If you're not familiar with differentiation" One can deduce this geometrically. The extremal point has to lie midway between the roots so it's the mean of the roots already found.
$endgroup$
– Winther
Nov 24 '18 at 11:03
$begingroup$
@Winther Well, I mentioned an alternative, which is a standard formula which can also be deduced without the use of differential calculus and is most often that not carried out in pre-calculus elementary algebra courses.
$endgroup$
– Rebellos
Nov 24 '18 at 11:05
$begingroup$
"If you're not familiar with differentiation" One can deduce this geometrically. The extremal point has to lie midway between the roots so it's the mean of the roots already found.
$endgroup$
– Winther
Nov 24 '18 at 11:03
$begingroup$
"If you're not familiar with differentiation" One can deduce this geometrically. The extremal point has to lie midway between the roots so it's the mean of the roots already found.
$endgroup$
– Winther
Nov 24 '18 at 11:03
$begingroup$
@Winther Well, I mentioned an alternative, which is a standard formula which can also be deduced without the use of differential calculus and is most often that not carried out in pre-calculus elementary algebra courses.
$endgroup$
– Rebellos
Nov 24 '18 at 11:05
$begingroup$
@Winther Well, I mentioned an alternative, which is a standard formula which can also be deduced without the use of differential calculus and is most often that not carried out in pre-calculus elementary algebra courses.
$endgroup$
– Rebellos
Nov 24 '18 at 11:05
add a comment |
$begingroup$
HINT
We have that for $ax^2+bx+c=0$
1) The solutions, or roots
- use that $x_1,x_2=frac{-bpm sqrt{b^2-4ac}}{2a}$
2) Coordinates of the vertex
- use that $x_V=-frac b{2a}$
3) equation of axis of symmetry
- see point $2$
For more details refer to Quadratic equation.
answered Nov 24 '18 at 9:24
gimusigimusi
92.9k84494
$endgroup$
add a comment |
$begingroup$
HINT
We have that for $ax^2+bx+c=0$
1) The solutions, or roots
- use that $x_1,x_2=frac{-bpm sqrt{b^2-4ac}}{2a}$
2) Coordinates of the vertex
- use that $x_V=-frac b{2a}$
3) equation of axis of symmetry
- see point $2$
For more details refer to Quadratic equation.
answered Nov 24 '18 at 9:24
gimusigimusi
92.9k84494
$endgroup$
add a comment |
$begingroup$
HINT
We have that for $ax^2+bx+c=0$
1) The solutions, or roots
- use that $x_1,x_2=frac{-bpm sqrt{b^2-4ac}}{2a}$
2) Coordinates of the vertex
- use that $x_V=-frac b{2a}$
3) equation of axis of symmetry
- see point $2$
For more details refer to Quadratic equation.
answered Nov 24 '18 at 9:24
gimusigimusi
92.9k84494
$endgroup$
HINT
We have that for $ax^2+bx+c=0$
1) The solutions, or roots
- use that $x_1,x_2=frac{-bpm sqrt{b^2-4ac}}{2a}$
2) Coordinates of the vertex
- use that $x_V=-frac b{2a}$
3) equation of axis of symmetry
- see point $2$
For more details refer to Quadratic equation.
answered Nov 24 '18 at 9:24
gimusigimusi
92.9k84494
answered Nov 24 '18 at 9:24
gimusigimusi
92.9k84494
answered Nov 24 '18 at 9:24
gimusigimusi
92.9k84494
answered Nov 24 '18 at 9:24
gimusigimusi
92.9k84494
92.9k84494
add a comment |
add a comment |
$begingroup$
this is what wolfram alpha gives: wolframalpha.com/input/?i=x%5E2-3x-4 and wolframalpha.com/input/?i=2x%5E2%2B7x%2B3
$endgroup$
– idea
Nov 24 '18 at 9:24
$begingroup$
@idea Not really helpful.
$endgroup$
– Rebellos
Nov 24 '18 at 9:24
$begingroup$
What did you try? Where are you stuck? Do you want roots of each? Do you want intersection points?
$endgroup$
– NoChance
Nov 24 '18 at 9:29
$begingroup$
Welcome (again) to Math.SE! To expand upon a comment your first question received ... The community here prefers/expects questions that include your thoughts, and/or an indication of where you got stuck. Such information helps answerers tailor responses to your skill level, without wasting anyone's time telling you what you already know. (It also helps convince people that you aren't simply trying to get them to do your homework for you.)
$endgroup$
– Blue
Nov 24 '18 at 9:32
$begingroup$
Your question mentions "each graph [...] shown", but shows no graph. You don't (yet) have enough reputation to post an image, but if you can link to an image (preferably hosted on a site such as Imgur.com), someone can add it to your question for you.
$endgroup$
– Blue
Nov 24 '18 at 9:34