Question

How do you find the vertex and intercepts for `f(x)=-9x^2 + 7x + 6`?

Answer 1

Vertex is at `(0.39 , 7.36)`, x-intercept is at `(1.29,0) and ( -0.52,0)` , y-intercept is at `(0,6)`

Explanation:

`f(x) = -9x^2 +7x +6 ; a= -9 , b=7 , c=6 ,` comparing with standard equation `f(x)=ax^2+bx+c`.

Vertex (x-ordinate) `=-b/(2a) = -7/-18=7/18 ~~0. 39(2dp)`
Vertex (y-ordinate) `f(x)=-9*7^2/18^2+7* 7/18 +6 ~~ 7.36(2dp)`

So Vertex is at `(0.39 , 7.36)`

y-intercept can be obtained by putting `x=0` in the equation , i.e `f(x)= -9*0 +7*0 +6 =6 :.` y-intercept is at `(0,6)`

x-intercept can be obtained by putting `f(x)=0` in the equation , i.e
`0 = -9x^2 +7x +6`
`:. x= - b/(2a) +- sqrt(b^2-4ac)/(2a)`
`:.x = -7/-18 +- sqrt (49 - 4 * -9 *6)/-18 =7/18 +- sqrt265/ -18`
`:. x ~~ - 0.52(2dp) , x ~~ 1.29 (2dp)`

x-intercept is at `(1.29,0) and ( -0.52,0)` graph{-9x^2+7x+6 [-20, 20, -10, 10]}[Ans]