Answer :

The answer is (0,-18) (2,0). Sorry it's late

Answer:

Refer the attached figure.

Step-by-step explanation:

Given : Quadratic function [tex]f(x)=2x^2+4x-16[/tex]

To find : The graph of the quadratic function using parabola tool?

Solution :

The given function [tex]f(x)=2x^2+4x-16[/tex]

First we find the vertex form of the equation

[tex]f(x)=2x^2+4x-16[/tex]

Where, a=2 ,b=4 , c=-16

Vertex is [tex]V=(\frac{-b}{2a},f(\frac{-b}{2a}))[/tex]

[tex]\frac{-b}{2a}=\frac{-4}{2(2)}=-1[/tex]

[tex]f(\frac{-b}{2a})=f(-1)=2(-1)^2+4(-1)-16=-18[/tex]

So, The vertex of the equation is (-1,-18)

Now, we find y- intercept by putting x=0 in the equation

[tex]y=2(0)^2+4(0)-16[/tex]

[tex]y=-16[/tex]

y- intercept (0,-16)

Now, we find x- intercept by putting y=0 in the equation

[tex]2x^2+4x-16=0[/tex]

[tex]x^2+2x-8=0[/tex]

[tex]x^2+4x-2x-8=0[/tex]

[tex]x(x+4)-2(x+4)=0[/tex]

[tex](x+4)(x-2)=0[/tex]

[tex]x=-4,2[/tex]

x- intercepts are (-4,0) and (2,0)

Placing all the points and plot a graph.

Refer the attached figure below.

${teks-lihat-gambar} tardymanchester

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