Find two points on the graph of the parabola other than the vertex and x-intercepts

Question

a(x)= x² +7x+12

Answer 1

f(x) = x^2 + 7x + 12 = (x+4)(x+3)

the x-intercepts are (-4,0) and (-3,0)

-b/(2a) = -7/2 = -3.5

f(-3.5) = (-3.5+4)(-3.5+3) = 0.5*-0.5 = -0.25 = -1/4 so the vertex is (-3.5, -0.25)

f(-5) = (-1)(-2) = 2 , so (-5,2) and f(-2) = (2)(1) =2, so (-2,2)

x y

-7 12

-5 2

-4 0

-3.5 -0.25

-3 0

-2 2

0 12

Answer 2

Great question!

One way to tackle this problem is by choosing some values for x.

For example, if you had a parabola that follows the equation y(x) = x²+2, you could pick a value for x.

Let's choose 2.

If x=2, what is the corresponding y value?

y(x) = 2²+2 = 4+2 = 6

So we know that (2, 6) is a point on the parabola y(x)=x²+2

Try this method with your parabola!

Hope this helps!