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# need help

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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

Sep 21, 2018

#1
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it would be nice to do all of them but its ok to just do one  :)

Sep 21, 2018
#2
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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.

Sep 21, 2018
#3
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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   Sep 22, 2018
#4
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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   Sep 22, 2018