October 28, 2015
Find the vertex of each parabola
y = 5(x-3)2 + 4
y = x2 - 4x + 7
y = 4x2 + 8x - 2
Find the x-intercepts and the y-intercept of
the function below.
y = 3x2 - 4x - 4
Write the quadratic equation in vertex form.
y = x2 - 6x + 5
complete the square
October 28, 2015
Sec. 2.1 Quadratic Functions
Quadratic Function
Let a, b, and c be real numbers with a≠0.
f(x) = ax + bx2 + c
The graph of a quadratic function is called a parabola
All parabolas have reflection symmetry about a line called
the axis of symmetry, which intersects the vertex
of the parabola.
October 28, 2015
For the function f(x)=ax , where a>0.2 Fill in the
characteristics below.
Domain
all real numbers
Range
y> 0
Intercept(s)
(0,0)
Decreasing on
the interval
(interval of x)
x<0
Increasing on
x> 0
Equation for axis
of symmetry
x=0
Coordinates of
Relative minimum
(0,0)
Symmetry
Even Function
Describe how the graph of each function
is related to the graph of y = x
a.) f(x) = .2x
2 shrink by factor of 5. Parabola will be wider
Vertical
b.) g(x) = 4x
2 stretch by factor of 4. Parabola will be narrower.
Vertical
c.) h(x) = -x +1
d.) k(x) = (x+2) -3
2
Reflection over the x-axis. Shift up 1.
2
Shift 2 left and 3 down.
2
October 28, 2015
Standard Form of a Quadratic Function
(Vertex)
f(x) = a(x - h) + k
2
where the vertex is (h, k)
Finding the vertex of a quadratic function.
First, write the function in standard form by
completing the square.
Isolate the x2
and x terms
Complete the
square by adding
(b/2)2 to both sides
Factor the perfect square
trinomial
Get y alone.
October 28, 2015
Complete the square
Complete the square
Leading coefficient not I
October 28, 2015
Complete the square
coeff of x2 must be 1
before you complete
the square
Finding the x-intercepts of a
quadratic function
Make y=0,
factor and
solve.
October 28, 2015
Find the x-intercepts for each function below.
f(x)=x2 +4x-12
0=(x+6)(x-2)
x = -6, x = 2
f(x)=2x2 -9x+4
0=(2x-1)(x-4)
x = 1/2, x = 4
Write the equation of a parabola in
standard form.
Find an equation, in standard form, for the
parabola with vertex (1,2) and
passing through the point (3, -6)
y = a(x-1)2+2
-6 = a(3-1)2+2
-6 = 4a + 2
-8 = 4a
-2 = a
Therefore,
f(x) = -2(x-1)2+2
substitute (1,2) for
(h, k) in the eq'n
and solve for a
October 28, 2015
Write the standard form of the equation of
the parabola with vertex (-4, 11) and
passing through the point (-6, 15).
y = a(x+4)2 +11
15 = a(-6 + 4)2 +11
15 = 4a + 11
4 = 4a
1=a
So, f(x) = (x+4)2 + 11
Finding Minimum and Maximum values
The vertex of a quadratic
function occurs at x = -b/2a
Let f(x) = -.0032x2 + x + 3 represent the
path of a baseball where f(x) is the height
of the baseball (in feet) and x is the
horizontal distance from home plate (in
feet). What is the maximum height
reached by the baseball?
V(1
.2
56
5,
8
2
1.1
5)
The maximum occurs
when x = -b/2a
or
x = -1/2(-.0032)
so
x = 156.25 feet.
Now find f(156.25).
f(156.25) = 81.125 feet
maximum height.
At what horizontal distance from home plate will the
ball hit the ground? (Hint: Find the x-intercepts)
October 28, 2015
Complete the square