How do you find the vertex and intercepts for #y = = 5x^2 + 4x - 3#?

1 Answer
Nghi N.
Jan 18, 2016

Vertex (-2/5, -19/5)

Explanation:

x-coordinate of vertex:
#x = -b/(2a) = -4/10 = -2/5#.
y-coordinate of vertex:
#y(-2/5) = 5(4/25) - 4(2/5) - 3 = 4/5 - 8/5 - 3 = -19/5#
To find y-intercept, make x = 0 --> y = -3
To find x-intercepts, make y = 0, and solve the quadratic equation:
5x^2 + 4x - 3 = 0
D = d^2 = b^2 - 4ac = 16 + 60 = 76 --> #d = +- sqrt76#
#x = -2/5 +- (sqrt76)/5#

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