+68 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
+118673 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.
+129852 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
+129852 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
