f(x)= x^2 +10x+31 in f(x) = a(x-h)^2 +k form

Question

given f(x)=x^2 +10x+31   I came up with vertex (-5,6) and the equation of                         f(x)= (X+5)^2 +6 Now how do I come up with points to graph?

Answer 1

good Job the vertex is (-5,6)

the axis of symetry is x = -5

use:f(x) = (x + 5)^2 + 6

set x = 0 to get the y-intercept

y-intercept = (5)^2 + 6 = 25 + 6 = 31, (0, 31)

There are no x-intercepts because the parabola faces upward and does not cross the x-axis.

there are three points go graph and good luck! Happy New Year!

I labeled the x-axis by 2 and the y-axis by 5

there are two points (-5,6), (0,31) to graph these 2 points remember it is a bowl so round the line.

to get the reflective point of (0,31) go five to the left on the x-axis and 31 up or (-10,31). Same here round the line for the bowl.

now plug in values to continue:

f(1) = (1 + 5)^2 + 6 = 36 + 6 = 42

f(2) = (2 + 5)^2 + 6 = 49 + 6 = 54

this gives you a good right side of the parabola

f(-11) = 42 reflective of f(1)

f(-12) = 54 reflective of f(2).

hope this helps and notice the changes.

 

 

Comments

thanks so much for reconfirming that I was on the right track  how do you graph -5+=sqrt6?