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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

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.

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 ?

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.