Create
Learn
Share

Quadratic Equations

rename
Updated 2009-03-12 00:08

Quadratic Equations

Quadratic Equations represent vertical parabolas and have the following forms:
y = a x2 + b x + c
y = a (x-h)2 + k
y = a (x-x1) (x-x2)

 

In the above, the following hold:
b = -2 a h = -a (x1 + x2).
The (x,y) coordinates for the vertex are (h,k):
h = -b/(2 a) and k = c - (b2)(4 a) = -(b2 - 4 a c)/(4 a), or
h = (x1 + x2)/2 and k = -a(x1 - x2)2/4.
If a>0, the parabola points up, and
the vertex (h,k) gives k as the minimum for y.
If a<0, the parabola points down, and
the vertex (h,k) gives k as the maximum for y.
The y-intercept has x=0 and the y value:
c = a h2 + k = a x1 x2.
The x-intercepts have y=0 and the x values:
[-b ± sqrt(b2 - 4 a c)]/(2 a),
h ± sqrt(-k/a), or
x1 and x2.
There is only one real x-intercept when:
b2 = 4 a c, or
k/a = 0.
There are no real x-intercepts when:
b2 < 4 a c, or
k/a > 0 (k and a are both positive or both negative).

 

Below are some examples in which the left column entries equal the right column entries.

You can try to memorize them, but it is better to use them as practice problems.

 

Left ColumnRight Column
y = 3 x2 - 12 x + 7y = 3 (x - 2)2 - 5
y = 3 x2 - 12 x + 7parabola's vertex (minimum) is (h,k)=(2,-5)
y = 3 x2 - 12 x + 7y = 3 (x - 2 + sqrt(5/3)) (x - 2 - sqrt(5/3))
y = 3 x2 - 12 x + 7x-intercepts have x = 2 ± sqrt(5/3) and y=0
y = 3 x2 - 12 x + 7y-intercept has x=0 and y=7
y = x2 - 6 x + 4y = (x - 3)2 - 5
y = x2 - 6 x + 4parabola's vertex (minimum) is (h,k)=(3,-5)
y = x2 - 6 x + 4y = (x - 3 + sqrt(5)) (x - 3 - sqrt(5))
y = x2 - 6 x + 4x-intercepts have x = 3 ± sqrt(3) and y=0
y = x2 - 6 x + 4y-intercept has x=0 and y=4
y = -x2 + 6 x - 5y = -(x - 3)2 + 4
y = -x2 + 6 x - 5parabola's vertex (maximum) is (h,k)=(3,4)
y = -x2 + 6 x - 5y = -(x - 1) (x - 5)
y = -x2 + 6 x - 5x-intercepts have x=1 or 5 and y=0
y = -x2 + 6 x - 5y-intercept has x=0 and y = -5
y = (x2/3) - (4 x)/3 - 11/3y = (1/3) (x - 2)2 - 5
y = (x2/3) - (4 x)/3 - 11/3parabola's vertex (minimum) is (h,k)=(2,-5)
y = (x2/3) - (4 x)/3 - 11/3y = (1/3) (x - 2 + sqrt(15)) (x - 2 - sqrt(15))
y = (x2/3) - (4 x)/3 - 11/3x-intercepts have x = 2 ± sqrt(15) and y=0
y = (x2/3) - (4 x)/3 - 11/3y-intercept has x=0 and y = -11/3
y = 12 x2 - 3parabola's vertex (minimum) is (h,k)=(0,-3)
y = 12 x2 - 3y = 12 (x + 1/2) (x - 1/2)
y = 12 x2 - 3x-intercepts have x = -1/2 or 1/2 and y=0
y = 12 x2 - 3y-intercept has x=0 and y = -3
memorize

 

Left ColumnRight Column
y = -2 x2 + 2 x + 24y = -2 (x - 1/2)2 + 49/2
y = -2 x2 + 2 x + 24parabola's vertex (maximum) is (h,k)=(1/2, 49/2)
y = -2 x2 + 2 x + 24y = -2 (x + 3) (x - 4)
y = -2 x2 + 2 x + 24x-intercepts have x = -3 or 4 and y=0
y = -2 x2 + 2 x + 24y-intercept has x=0 and y=24
y = 3 x2 - 21 x + 30y = 3 (x - 7/2)2 - 27/4
y = 3 x2 - 21 x + 30parabola's vertex (minimum) is (h,k)=(7/2, -27/4)
y = 3 x2 - 21 x + 30y = 3 (x - 2) (x - 5)
y = 3 x2 - 21 x + 30x-intercepts have x=2 or 5 and y=0
y = 3 x2 - 21 x + 30y-intercept has x=0 and y=30
y = -5 x2 - 30 x - 45y = -5 (x + 3)2
y = -5 x2 - 30 x - 45parabola's vertex (maximum) is (h,k)=(-3,0)
y = -5 x2 - 30 x - 45x-intercept has x = -3 and y=0
y = -5 x2 - 30 x - 45y-intercept has x=0 and y= -45
y = 7 x2 - 7parabola's vertex (minimum) is (h,k)=(0, -7)
y = 7 x2 - 7y = 7 (x + 1) (x - 1)
y = 7 x2 - 7x-intercepts have x = -1 or 1 and y=0
y = 7 x2 - 7y-intercept has x=0 and y = -7
y = (x2/2) + 4 x + 15/2y = (1/2) (x + 4)2 - 1/2
y = (x2/2) + 4 x + 15/2parabola's vertex (minimum) is (h,k)=(-4, -1/2)
y = (x2/2) + 4 x + 15/2y = (1/2) (x + 3) (x + 5)
y = (x2/2) + 4 x + 15/2x-intercepts have x = -3 or -5 and y=0
y = (x2/2) + 4 x + 15/2y-intercept has x=0 and y=15/2
y = (x2/2) + 4 x + 8y = (1/2) (x + 4)2
y = (x2/2) + 4 x + 8parabola's vertex (minimum) is (h,k)=(-4,0)
y = (x2/2) + 4 x + 8x-intercept has x = -4 and y=0
y = (x2/2) + 4 x + 8y-intercept has x=0 and y=8
memorize

 

Left ColumnRight Column
y = -5 x2 + 5 x + 10y = -5 (x - 1/2)2 + 45/4
y = -5 x2 + 5 x + 10parabola's vertex (maximum) is (h,k)=(1/2, 45/4)
y = -5 x2 + 5 x + 10y = -5 (x + 1) (x - 2)
y = -5 x2 + 5 x + 10x-intercepts have x = -1 or 2 and y=0
y = -5 x2 + 5 x + 10y-intercept has x=0 and y=10
y = -4 x2 + 3 x + 7y = -4 (x - 3/8)2 + 121/16
y = -4 x2 + 3 x + 7parabola's vertex (maximum) is (h,k)=(3/8, 121/16)
y = -4 x2 + 3 x + 7y = -4 (x - 7/4) (x + 1)
y = -4 x2 + 3 x + 7x-intercepts have x=7/4 or -1 and y=0
y = -4 x2 + 3 x + 7y-intercept has x=0 and y=7
y = -4 x2 - 3 x + 7y = -4 (x + 3/8)2 + 121/16
y = -4 x2 - 3 x + 7parabola's vertex (maximum) is (h,k)=(-3/8, 121/16)
y = -4 x2 - 3 x + 7y = -4 (x + 7/4) (x - 1)
y = -4 x2 - 3 x + 7x-intercepts have x = -7/4 or 1 and y=0
y = -4 x2 - 3 x + 7y-intercept has x=0 and y=7
y = 2 x2 + 3 x - 5y = 2 (x + 3/4)2 - 49/8
y = 2 x2 + 3 x - 5parabola's vertex (minimum) is (h,k)=(-3/4, -49/8)
y = 2 x2 + 3 x - 5y = 2 (x + 5/2) (x - 1)
y = 2 x2 + 3 x - 5x-intercepts have x = -5/2 or 1 and y=0
y = 2 x2 + 3 x - 5y-intercept has x=0 and y = -5
y = 2 x2 + 3 x + 5y = 2 (x + 3/4)2 + 31/8
y = 2 x2 + 3 x + 5parabola's vertex (minimum) is (h,k)=(-3/4, 31/8)
y = 2 x2 + 3 x + 5y = 2 (x + 3/4 + (i/4) sqrt(31)) (x + 3/4 - (i/4) sqrt(31))
y = 2 x2 + 3 x + 5there are no intercepts on the real x axis
y = 2 x2 + 3 x + 5x-intercepts have x = -3/4 ± (i/4) sqrt(31) and y=0
y = 2 x2 + 3 x + 5y-intercept has x=0 and y=5
memorize