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

Question

Hazel Anglin · Accepted Answer

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

Abigail Allen · Answer

undefined To find the vertex of the quadratic function $ f(x) = -9x^2 + 7x + 6 $, use the formula $ x = -\frac{b}{2a} $. Then, substitute this $ x $ value into the function to find the corresponding $ y $ value.

To find the $ x $-intercepts, set $ f(x) = 0 $ and solve for $ x $ using the quadratic formula.

To find the $ y $-intercept, substitute $ x = 0 $ into the function.