How do you find the vertex and intercepts for #f(x) = 10x^2 - 5x + 12#?

1 Answer
Jun 28, 2018

Vertex (#1/4, 91/8)#
No x-intercepts

Explanation:

#f(x) = 10x^2 - 5x + 12#
x - coordinate of vertex:
#x = -b/(2a) = 5/20 = 1/4#
y-coordinate of vertex:
#f(1/4) = 10(1/16)- 5(1/4) + 12 = 5/8 - 10/8 + 96/8 = 91/8#
Vertex #(1/4, 91/8)#
y-intercept when x = 0, --> f(0) = 12
x- intercepts when f(x) = 0. Solve the quadratic equation:
#f(x) = 10x^2 - 5x + 12 = 0#
#D = d^2 = b^2 - 4ac = 25 - 480 = -455 < 0.#
There are no x-intercepts. The graph of f(x) is an upward parabola (a > 0), that completely stays above the x-axis.