How do you find the vertex and the intercepts for #q(x) = 1/2x² + 2x + 5#?

1 Answer
Feb 1, 2017

vertex (-2, 3)

Explanation:

#q(x) = x^2/2 + 2x + 5#
x-cordinate of vertex:
#x = -b/(2a) = - 2/1 = - 2#
y-coordinate of vertex:
#q(-2) = 2 - 4 + 5 = 3#
Vertex (-2, 3)
To find y-intercept, make x = 0 --> y = 5
To find x-intercepts, make y = 0, and solve the quadratic equation
#x^2/2 + 2x + 5 = 0#
#D = b^2 - 4ac = 4 - 10 = - 6 < 0.#
Since D < 0, there are no real roots, meaning no x-intercepts.
The parabola graph is completely above the x-axis.
graph{x^2/2 + 2x + 5 [-20, 20, -10, 10]}