How do you find the vertex and axis of symmetry, and then graph the parabola given by: #y = -2x^2 + 10x - 1#?

1 Answer
Oct 16, 2015

Vetex #(5/2, (-23)/2)#

Axis of symmetry - #x=5/2#

Explanation:

#-2x^2+10x-1#

Find the vertex

#x=(-b)/(2a)=(-10)/(2 xx(-2))=(-10)/(-4)=5/2#

#y=-2(5/2)^2+10(5/2)-1=-2(25/4)+25-1#
#y=-(25/2)+25-1 = (-25+50-2)/2=(-23)/2#

Vetex #(5/2, (-23)/2)#

Axis of symmetry -

#x=5/2#

Since co-efficent of #x^2# is negative, the curve opens downwards. The function has a maximum at the vertex.

Take a few points less than #x=2.5# and a few points greater.
Calculate their corresponding #y# values.

x y
0 -1
1 7
2 11
2.5 11.5
3 11
4 7
5 -1
Plot the points and join them with a smooth curve.

graph{-2x^2+10x-1 [-32.48, 32.48, -16.24, 16.22]}