How do you find the vertex and the intercepts for #y = -2x^2 + 13x - 6#?

1 Answer
Apr 7, 2016

#Vertex(13/4, 121/8)#

Explanation:

x-coordinate of vertex:
#x = -b/(2a) = -13/-4 = 13/4#
y-coordinate of vertex:
#y(13/4) = -2(169/16) + 13(13/4) - 6 = 169/8 - 6 = 121/8#
Vertex #(13/4, 121/8)#
To find y-intercept, make x = 0 --> y = -6.
To find x-intercepts, make y = 0 and solve the quadratic equation
#-2x^2 + 13x - 6 = 0#
#D = d^2 = b^2 - 4ac = 169 - 48 = 121# --> # d = +- 11#
There are 2 real roots (x-intercepts):
#x = - b/(2a) +- d/(2a) = 13/4 +- 11/4#
graph{-2x^2 + 13x - 6 [-40, 40, -20, 20]}