Answer:
the x intercepts are [tex](1-\sqrt{5} ,0) , (\:1+\sqrt{5},0 )[/tex]
Explanation:
vertex (1,20) and y-intercept (0,16)
equation used f(x) = a(x - h)² + k where (h,k) is vertex.
16 = a(0 - 1)² + 20
16 = a + 20
a = -4
f(x) = a(x - h)² + k ......this is vertex to find equation of parabola.
f(x) = -4(x-1)² + 20
f(x) = -4(x-1)² + 20
f(x) = -4x²-8x-4+20
f(x) = -4x²+8x+16 .....if simplified.
To find x intercepts, y must be 0,
4x²+8x+16 =0
[tex]x_{1,\:2}=\frac{-8\pm \sqrt{8^2-4\left(-4\right)\cdot \:16}}{2\left(-4\right)}[/tex]
[tex]x_1=\frac{-8+8\sqrt{5}}{2\left(-4\right)},\:x_2=\frac{-8-8\sqrt{5}}{2\left(-4\right)}[/tex]
[tex]x=1-\sqrt{5},\:x=1+\sqrt{5}[/tex]
So, the x-intercepts of the parabola vertex (1,20) and y-intercept (0,16) are [tex](1-\sqrt{5} ,0) , (\:1+\sqrt{5},0 )[/tex]
