Graph quadratic functions in vertex form

Question

Graph the function  f(x)= -1/4x (x+8)^2 +1

I was able to find the point (-8, 1)

Instruction: Now look for another point on the parabola with integer or half-integer coordinates. One such point is 

(-6, 0)

My question is how to how figure the point (-6, 0) ?

Answer 1

The fact that you found (-1,8) means that you understand one of the biggest benefits of vertex form, it easily allows you to find the vertex. From there, they want you to find other points on the parabola. I would choose to use integer and not half integer if given the choice, because I wouldn't want to work with fractions. You can pick the x coordinates that are close to the vertex and then use the equation to solve for the y values.

If you try -6

f(x)= -1/4 (-6+8)² +1

f(x)=0

So one point on the parabola is (-6,0)

If you try -4

f(x)= -1/4 (-4+8)² +1

f(x)=-3

So another point on the parabola is (-4,-3)

Parabola are symmetric. So once you find these points, you will also have the points on the other side of the vertex, they have the same y value.

(-10,0) (You could do the math on these to prove it to yourself.)

(-12,-3)

Now that you have 5 points, you can make a nice graph.

Answer 2

0 = -1/4(-6 + 8)² + 1

0 = -1/4(2)² + 1

0 = -1/4(4) + 1

0 = -1 + 1

0 = 0