Question

Consider h(x) = 3x^2 +2x - 8

Identify its vertex and y-intercept.
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Algebra 2

Answer 1

Vertex:
`(-(1)/(3),-(25)/(3))`

Y-intercept:
`(0,-8)`

Explanation:

The x-coordinate of the vertex would be
`x=(-b)/(2a)` and the y-coordinate of the vertex is whatever the output is given the value of x.

Therefore, the x-coordinate of the vertex is
`x=(-b)/(2a)=(-2)/(2(3))=(-2)/(6)=-(1)/(3)`

This means the y-coordinate of the vertex is
`y=3x^2+2x-8=3(-(1)/(3))^2+2(-(1)/(3))-8=3((1)/(9))-(2)/(3)-8=(1)/(3)-(2)/(3)-8=-(1)/(3)-8=-(1)/(3)-(24)/(3)=-(25)/(3)`

So, the vertex is
`(-(1)/(3),-(25)/(3))`

The y-intercept of a function is the y-value at which x=0, or the y-value when the function crosses the y-axis. Therefore, if we plug x=0 into the function, we see that
`h(0)=3(0)^2+2(0)-8=0+0-8=-8`, so our y-intercept is -8 or (0,-8).

I attached a graph below to help you visualize the vertex and y-intercept given the function.