Algebra I. Finding vertex in y= (x-4) (x+2).?

Question

Hello.
I'm in middle school and I have Algebra I.

Find the x-intercepts and the vertex of the graph of the function. Then sketch the graph of the function.
Problem: y= (x-4) (x+2)

I have found out the x-intercepts, which would be x= 4 or x= -2.
I needed help on finding the vertex, the graphing I know though.
Thank you.

Answer 1

You know two points on the graph; (4,0) and (-2,0), for those are the x intercepts you stated above.
Since Parabolas are always symmetrical, you know that the x coordinate of the vertex must be exactly half way between these two coordinates(since they have the same y value), somewhere along the line x=1
plug 1 into the equation, and see what value you get for y:
y = (1-4)(1+2)
y = -3(3)
y = -9
vertex = (1,-9)

Answer 2

f(x) = x ² - 2x - 8

f ` (x) = 2x - 2 = 0 for turning point

x = 1 for turning point

f " (x) = 2 is +ve so MIN turning point

f (1) = - 9

(1 , - 9) is a MIN turning point (vertex)

f(0) = - 8 gives point (0,- 8)

Cuts x axis at x = 4 and x = - 2

Hence graph.

Answer 3

(a million, -9) amplify the equation. so y= x^2 -2x -8. then you certainly desire to end the sq.: y=(x^2 -2x) -8 y=(x^2 -2x +a million) -8 -a million y=(x-a million)^2 -9 the kind for the parabola is y= a(x-h)^2 + ok (h,ok) is the vertex word: x^2 is x-squared