Question

Which is the graph of the function f(x) = one-halfx2 + 2x – 6?

Answer 1

Answer:

The answer is A

Step-by-step explanation:

trust me, i just did it and got it right, don't listen to that dumb fool who did that dumb answer

Answer 2

Answer:

On a coordinate plane, a parabola opens up. It goes through (negative 6, 0), has a vertex at (negative 2, negative 8), and goes through (2, 0)

Step-by-step explanation:

step 1

Find the vertex

we have

$f(x)=\frac{1}{2}x^{2}+2x-6$

This is a vertical parabola open upward (because the leading coefficient is positive)

The vertex represent a minimum

Convert the equation in vertex form

Factor 1/2

$f(x)=\frac{1}{2}(x^{2}+4x)-6$

Complete the square

$f(x)=\frac{1}{2}(x^{2}+4x+4)-6-2$

$f(x)=\frac{1}{2}(x^{2}+4x+4)-8$

Rewrite as perfect squares

$f(x)=\frac{1}{2}(x+2)^{2}-8$

The vertex is the point (-2,-8)

step 2

Find the x-intercepts

we know that

The x-intercepts are the values of x when the value of y is equal to zero

so

For f(x)=0

$\frac{1}{2}(x+2)^{2}-8=0$

$\frac{1}{2}(x+2)^{2}=8$

$(x+2)^{2}=16$

square root both sides

$x+2=(+/-)4$

$x=-2(+/-)4$

$x=-2(+)4=2$

$x=-2(-)4=-6$

so

The x-intercepts are the points (-6,0) and (2,0)

therefore

On a coordinate plane, a parabola opens up. It goes through (negative 6, 0), has a vertex at (negative 2, negative 8), and goes through (2, 0)