One x-intercept for a parabola is at the point(2,0). Use the quadratic formula to find theother x-intercept for the parabola defined bythis equation:y = 4x2 - 4x – 8

Accepted Solution

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:[tex]\text{ If } y=ax^2+bx+c \text{ then the } x-\text{intercepts are } (\frac{-b \pm \sqrt{b^2-4ac}}{2a},0)[/tex].Let's start:Compare the following equations to determine the values for [tex]a,b, \text{ and }c [/tex]:[tex]y=ax^2+bx+c[/tex][tex]y=4x^2-4x-8[/tex]So [tex]a=4[/tex][tex]b=-4[/tex][tex]c=-8[/tex]We are now ready to enter into our formula:[tex]x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}[/tex][tex]x=\frac{4 \pm \sqrt{(-4)^2-4(4)(-8)}}{2(4)}[/tex][tex]x=\frac{4 \pm \sqrt{16+16(8)}}{8}[/tex][tex]x=\frac{4 \pm \sqrt{16(1+8)}}{8}[/tex][tex]x=\frac{4 \pm \sqrt{16}\sqrt{1+8}}{8}[/tex][tex]x=\frac{4 \pm 4\sqrt{9}}{8}[/tex][tex]x=\frac{ 4 \pm 4(3)}{8}[/tex][tex]x=\frac{4 \pm 12}{8}[/tex][tex]x=\frac{4(1\pm 3)}{8}[/tex][tex]x=\frac{1(1\pm 3)}{2}[/tex][tex]x=\frac{1 \pm 3}{2}[/tex][tex]x=\frac{1+3}{2} \text{ or } \frac{1-3}{2}[/tex][tex]x=\frac{4}{2} \text{ or } \frac{-2}{2}[/tex][tex]x=2 \text{ or } -1[/tex]So the x-intercepts are (2,0) and (-1,0).(2,0) was already given so (-1,0) is the other one.