Consider the following function

Question

Find two points of the parabola other than the vertex and x intercepts q(x)=x²+7x+12

Answer 1

q(x)= y = x^2 +7x + 12

other than the vertex and x intercepts is the y intercept found by setting x=0

that's easy, y= 12

y= 0^2 + 7(0) +12

y = 0 +0 +12

y = 12

that's the point (0,12)

let x = 1

then

y = 1+7+12 = 20

that's the point (1, 20)

they are 2 points other than the vertex or x intercepts

(0,12) and (1,20)

the vertex is the minimum point on the upward opening parabola

where the derivative = 0

y' = 2x +7 = 0

x =-7/2 = -3.5

y = (-3.5)^2 +7(-3.5) +12

= 12.25 -24.5 +12

= -.25

that's the point (-3.5, -.25)

you might have wanted to do that to check that your other 2 points are not the same as the vertex

y=x^2 +7x + 12

= x^2 +7x + (7/2)^2 + 12 - (7/2)^2

= (x+7/2)^2 +12 - 49/4

= (x+3.5)^2 - .25

= (x-h)^2 +k where (h,k) = vertex = (-3.5, -.25)