Question

What are the vertex and axis of symmetry of the parabola y=3x^2-24x+38?

Answer 1

see explanation

Explanation:

given a parabola in standard form : y = ax² + bx + c : a ≠ 0

Then the x- coordinate of the vertex which is also the axis of symmetry is

`x_(vertex)` = -
`(b)/(2a)`

y = 3x² - 24x + 38 is in standard form

with a = 3, b = - 24 and c = 38, hence

`x_(vertex)` = -
`(-24)/(6)` = 4

Substitute x = 4 into the equation for corresponding y- coordinate

y = 3(4)² - 24(4) + 38 = 48 - 96 + 38 = - 10

vertex = (4, - 10)

equation of axis of symmetry is x = 4