2.1
Quadratic Functions
Use the graphing calculator
and graph y = x2
y = 2x2
1 2
y= x
3
The quadratic function f(x) = a(x – h)2 + k
is said to be in standard form. Vertex (h,k)
If a > 0, opens up. If a < 0, opens down.
Ex.
Write the quadratic function in standard
form, find the vertex, identify the x-int.’s
and sketch. f(x) = 2x2 + 8x + 7
f(x) = 2x2 + 8x + 7
Factor out a 2 from the x’s
f(x) = 2(x2 + 4x + 4 ) + 7 - 8 Complete the square
f(x) = 2(x + 2)2 - 1
V(-2 , -1 )
Up or Down
Take the original and set = 0 to find the x-int.
0 = 2x2 + 8x + 7
Use quad. formula on calc.
x = -1.2928, -2.7071
Now sketch it.
(-2,-1)
Ex.
Sketch the graph of f(x) = -x2 + 6x - 8
f(x) = -x2 + 6x - 8
Factor -1 out of the x terms
f(x) = -(x2 – 6x + 9 ) - 8 + 9
f(x) = -(x – 3)2 + 1
Complete the square
V(3, 1)
Down
To find a vertex of the quadratic f(x) = ax2 + bx + c,
(without putting the equation into standard form),
evaluate by letting
!b
x=
2a
Ex.
P = .0014x2 - .1529x + 5.855
b
" .1529
x="
="
! 54.6
2a
2(.0014)
y = 1.68
Find the the equations of a parabola that cups up
and down given the x-intercepts (1, 0) and (-3, 0).
First, write the x-intercepts as factors.
y = (x - 1)(x + 3)
The parabola that cups up is y = x2 + 2x - 3
The parabola that cups down requires a negative
in front of the quadratic.
y = -(x2 + 2x - 3)
or y = -x2 - 2x + 3
Given the vertex and point on a given parabola, find
the standard equation of the parabola.
V(4, -1); point (2, 3)
Fill in the points in the standard parabola form
y = a(x - h)2 + k and solve for a.
3 = a(2 - 4)2 + (-1)
3 = 4a -1
4 = 4a
4=a
The answer is
y = 4(x - 4)2 - 1