How do you find the vertex of the parabola: y=-5x^2+10x+3y=−5x2+10x+3?

1 Answer
jigglypuff314
Aug 10, 2015

The vertex is (1,8)(1,8)

Explanation:

The x point of the vertex (x, y)(x,y) is located on the parabola's Axis of Symmetry.
~
The Axis of Symmetry of a Quadratic Equation
can be represented by x=-b/{2a}x=−b2a
when given the quadratic equation y=ax^2+bx+cy=ax2+bx+c
~
In this case, given that y=-5x^2+10x+3y=−5x2+10x+3
we can see that a=-5a=−5 and b=10b=10

plugging this into x=-b/{2a}x=−b2a
will get us: x=-10/{2*(-5)}x=−102⋅(−5)
which simplifies to x=1x=1
~
Now that we know the x value of the vertex point, we can use it to find the y value of the point!
Plugging x=1x=1 back into y=-5x^2+10x+3y=−5x2+10x+3
we will get: y=-5+10+3y=−5+10+3
which simplifies to: y=8y=8
~
so we have x=1x=1 and y=8y=8
for the vertex point of (x,y)(x,y)
therefore the vertex is (1,8)(1,8)

Related questions
See all questions in Quadratic Functions and Their Graphs
Impact of this question
7642 views around the world
You can reuse this answer
Creative Commons License