Question

One x-intercept for a parabola is at the point

Answer 1

Answer:

(2,0) was already given so (-1,0) is the other one.

Step-by-step explanation:

So we are asked to use the quadratic formula.

To find the x-intercepts (if they exist) is use:

$\text{ If } y=ax^2+bx+c \text{ then the } x-\text{intercepts are } (\frac{-b \pm \sqrt{b^2-4ac}}{2a},0)$.

Let's start:

Compare the following equations to determine the values for $a,b, \text{ and }c$:

$y=ax^2+bx+c$

$y=4x^2-4x-8$

So

$a=4$

$b=-4$

$c=-8$

We are now ready to enter into our formula:

$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$

$x=\frac{4 \pm \sqrt{(-4)^2-4(4)(-8)}}{2(4)}$

$x=\frac{4 \pm \sqrt{16+16(8)}}{8}$

$x=\frac{4 \pm \sqrt{16(1+8)}}{8}$

$x=\frac{4 \pm \sqrt{16}\sqrt{1+8}}{8}$

$x=\frac{4 \pm 4\sqrt{9}}{8}$

$x=\frac{ 4 \pm 4(3)}{8}$

$x=\frac{4 \pm 12}{8}$

$x=\frac{4(1\pm 3)}{8}$

$x=\frac{1(1\pm 3)}{2}$

$x=\frac{1 \pm 3}{2}$

$x=\frac{1+3}{2} \text{ or } \frac{1-3}{2}$

$x=\frac{4}{2} \text{ or } \frac{-2}{2}$

$x=2 \text{ or } -1$

So the x-intercepts are (2,0) and (-1,0).

(2,0) was already given so (-1,0) is the other one.