Precalculus Examples
(0,1)(0,1) , (1,0)(1,0)
Step 1
The general equation of a parabola with vertex (h,k)(h,k) is y=a(x-h)2+ky=a(x−h)2+k. In this case we have (0,1)(0,1) as the vertex (h,k)(h,k) and (1,0)(1,0) is a point (x,y)(x,y) on the parabola. To find aa, substitute the two points in y=a(x-h)2+ky=a(x−h)2+k.
0=a(1-(0))2+10=a(1−(0))2+1
Step 2
Step 2.1
Rewrite the equation as a(1-(0))2+1=0a(1−(0))2+1=0.
a(1-(0))2+1=0a(1−(0))2+1=0
Step 2.2
Simplify each term.
Step 2.2.1
Subtract 00 from 11.
a⋅12+1=0a⋅12+1=0
Step 2.2.2
One to any power is one.
a⋅1+1=0a⋅1+1=0
Step 2.2.3
Multiply aa by 11.
a+1=0a+1=0
a+1=0a+1=0
Step 2.3
Subtract 11 from both sides of the equation.
a=-1a=−1
a=-1
Step 3
Using y=a(x-h)2+k, the general equation of the parabola with the vertex (0,1) and a=-1 is y=(-1)(x-(0))2+1.
y=(-1)(x-(0))2+1
Step 4
Step 4.1
Remove parentheses.
y=(-1)(x-(0))2+1
Step 4.2
Multiply -1 by (x-(0))2.
y=-1(x-(0))2+1
Step 4.3
Remove parentheses.
y=(-1)(x-(0))2+1
Step 4.4
Simplify each term.
Step 4.4.1
Subtract 0 from x.
y=-1x2+1
Step 4.4.2
Rewrite -1x2 as -x2.
y=-x2+1
y=-x2+1
y=-x2+1
Step 5
The standard form and vertex form are as follows.
Standard Form: y=-x2+1
Vertex Form: y=(-1)(x-(0))2+1
Step 6
Simplify the standard form.
Standard Form: y=-x2+1
Vertex Form: y=-1x2+1
Step 7