Question 734089
Complete the square in graphing form: y = x^2 - 7x + 5?
.
The idea to keep in mind is that the "vertex form" of a quadratic is:
y = a(x – h)^2 + k
where
(h,k)is the vertex
.
y = x^2 - 7x + 5
since the y is already in the correct position, complete the square ONLY on the right side:
y = x^2 - 7x + 5
group x's:
y = (x^2 - 7x) + 5
add (b/2)^2 to the inside of the () and balance by subtracting on the outside:
y = (x^2 - 7x+12.25) + 5-12.25
y = (x-3.5)^2-7.25
from "y = a(x – h)^2 + k"
we see that the vertex is:
(3.5, -7.25)
.
First point on the graph (i.e., the lowest point on the graph)
.
To find y-intercepts, set x=0 and solve for y:
y = (x-3.5)^2-7.25
y = (0-3.5)^2-7.25
y = (-3.5)^2-7.25
y = 12.25-7.25
y = 4.75
Second point is at (0, 4.75)
.
To find x-intercepts, set y=0 and solve for x:
0 = (x-3.5)^2-7.25
7.25 = (x-3.5)^2
take the square root of both sides:
sqrt(7.25) = x-3.5
+-sqrt(7.25)+3.5 = x
6.24 = x
and
0.81 = x
.
Third point is at (6.24,0)
Fourth point is at (.81,0)