1.Find the vertex of the graph of the equation x-y^2+6y=8.

(Enter your submission as an ordered pair. For example: "(0,1)" but without the quotes.)

2.Find the vertex of the graph of the equation y=3x^2-6x+7.

3.The graph of the equation y=ax^2+bx+c, where a,b, and c are constants, is a parabola with axis of symmetry x=-3.

Find b/a

isthebest123 Sep 21, 2018

#2**+1 **

The middle one should be easy. I'm going to give hints.

How do you find the roots?

you have to simultaneously solve y=0 with that parabola. That will give the points where the p[arabola cuts the x axis.

The vertex is on the axis of symmetry and the axis of symm is half way between the roots.

Once you have the x value you can find the y with substitution.

the roots are

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\) seperate those 2 out and you will see that they are symmetrical about a very easy to find point.

Melody Sep 21, 2018

#3**+1 **

1) x-y^2+6y=8

Rearrange as

(x - 8) = y^2 - 6y complete the square on the right side

x - 8 + 9 = y^2 - 6y + 9 factor the right side and simplify the left

(x + 1 ) = ( y - 3)^2

In the form (x - h) = (y - k)^2....the vertex is (h , k) = (-1, 3)

See the graph, here : https://www.desmos.com/calculator/h9ee3oueph

CPhill Sep 22, 2018

#4**+1 **

3.The graph of the equation y=ax^2+bx+c, where a,b, and c are constants, is a parabola with axis of symmetry x = -3.

Find b/a

This parabola is vertical....the axis of symmetry is x = -3, which means that the x coordinate of the vertex is - 3

The x coordinate of the vertex is given by - b /a

So

-b/a = -3 multiply both sides by -1

b / a = 3

CPhill Sep 22, 2018