Graph the quadratic function

Question

Find the x-intercepts intercept and y intercept. Next find the vertex. Write the coordinates in decimal form. Then graph the function by hand. y=x^2-12x+32

Answer 1

vertex is when x=6

you can find that a couple ways. Simplest is take the derivative, set =0, solve for x

f'(x) = 2x -12 = 0

2x =12

x= 12/2 = 6

y intercept = 32, the constant term in f(x0

x intercepts = 6+ or -2 = 8 or 4

use the quadratic formula on x^2 -6x +32

vertex has y coordinate of 6^2 -12(6) +32 = -4

vertex = (6,-4) = minimum point

it's an upward opening parabola. You have 3 points, the vertex and 2 intercepts. Connect them with a smooth parabolic curve axis of symmetry is x=6, so another point on the parabola graph is (12, 32) symmetric to (0,32)