Question

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

Answer 1

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]}