How do you find the vertex and intercepts for #y = 8x + 5x^2 – 3#?

1 Answer
Jul 12, 2016

Vertex #(-4/5, -31/5)#

Explanation:

#y = 5x^2 + 8x - 3#
x-coordinate of vertex:
#x = -b/(2a) = -8/10 = -4/5#
y-coordinate of vertex:
#y(-4/5) = 5(16/25) - 32/5 - 3 = 16/5 - 32/5 - 3 = - 31/5#
Vertex #(-4/5, -31/5)#
To find y-intercept, make x = 0 --> y = -3
To find 2 x-intercepts, make y = 0 and solve the quadratic equation #y = 5x^2 + 8x - 3 = 0#
#D = d^2 = b^2 - 4ac = 64 + 60 = 124# --> #d = +- 2sqrt31#
There are 2 x-intercepts (2 real roots):
#x = -b/(2a) +- d/(2a) = -8/10 +- (2sqrt31)/10 = -4/5 +- sqrt31/5#
#x = (-4 +- sqrt31)/5#