Question

How do you find the vertex and the intercepts for `f(x) = 3 x^2 − 14 x − 24`?

Answer 1

Vertex `(14/6, -121/3)`

Y - intercept `(0, -24)`
One of the x-intercept `(6,0)`

Another x-intercept `(-4/3, 0)`

Explanation:

Given -

`y=3x^2-14x-24`

Find the Vertex

x-coordinate of the vertex

`x=(-b)/(2a)=(-(-14))/(2xx3)=14/6`

y-coordinate of the vertex

`y=3(14/6)^2-14(14/6)-24=-121/3`

Vertex `(14/6, -121/3)`

Y - intercept

At `x=0: y=3(0)^2 -14(0)-24=-24`

`(0, -24)`

X - intercepts

At `y=0`

`3x^2-14x-24=0`

`x=((-b)+-sqrt(b^2-(4ac)))/(2a)=(-(-14)+-sqrt((-14)^2-(4xx3xx(-24))))/(2xx3)`

`x=(b+-sqrt(196 -(-288)))/6=(14+-22)/6`

`x=(14+22)/6=6`

One of the x-intercept `(6,0)`

`x=(14-22)/6=(-8)/6=(-4)/3`

Another x-intercept `(-4/3, 0)`