Transcript 4.10 PowerPoint

Warm-up Quiz
1. Graph the inequality.
y > – x2 + 4x – 3
2. Graph the system of inequalities.
y ≤ – x2 + 4
y > x2 – 2x – 3
What are the 3 forms of quadratic equations for graphing?
1. Standard
2. Intercept
3. Vertex
Today’s Objective (Section 4.10): To be able to write a quadratic equation when
given certain information.
In previous sections we were given the equation and asked to
graph it, now we will be working backwards!!
Keep It So Simple . . . .
Use the most convenient form!!
Graphing y =
2
x
Standard
Intercept
y = ax2+bx+c
y = a (x - p) (x - q) y= a (x - h)2+k
axis:
x
b
2a
axis:
x
Vertex
pq
2
axis: x = h
vertex ( x , y )
vertex ( x , y )
vertex ( h , k )
C is y-intercept
x-intercept = p , q
a>0 U shaped,
Minimum
a>0 U shaped,
Minimum
a>0 U shaped,
Minimum
a<0 ∩ shaped,
Maximum
a<0 ∩ shaped,
Maximum
a<0 ∩ shaped,
Maximum
Write a quadratic function for the parabola shown.
Use the most convenient form!!
Use vertex form because the
vertex is given.
y = a(x – h)2 + k
Vertex form
y = a(x – 1)2 – 2
Substitute 1 for h and –2 for k.
Use the other given point, (3, 2), to find a.
Substitute 3 for x and 2 for y.
2 = a(3 – 1)2 – 2
Simplify coefficient of a.
2 = 4a – 2
Solve for a.
1=a
So . . . A quadratic function for the graph of the parabola is
y = (x – 1)2 – 2
Write a quadratic function for the parabola shown.
Use the most convenient form!!
Use intercept form because the
x-intercepts are given.
y = a(x – p)(x – q)
Intercept form
y = a(x + 1)(x – 4)
Substitute –1 for p and 4 for q.
Use the other given point, (3, 2), to find a.
2 = a(3 + 1)(3 – 4)
2 = – 4a
– 1 =a
2
So . . .
Substitute 3 for x and 2 for y.
Simplify coefficient of a.
Solve for a.
y=
–
1
2
(x + 1)(x – 4) .
Write a quadratic function in standard form for the
parabola that passes through the points (–1, –3),
(0, – 4), and (2, 6).
STEP 1
Substitute the coordinates of each point into
y = ax2 + bx + c to obtain the system of three linear
equations shown below.
–3 = a(–1)2 + b(–1) + c
–3 = a – b + c
Equation 1
–4 = a(0)2 + b(0) + c
–4=c
Equation 2
6 = a(2)2 + b(2) + c
6 = 4a + 2b + c
Equation 3
STEP 2
Solve the three equations to get:
a=2
b=1
c=–4
So . . . The equation in standard form for the
three points given is:
y = 2x2 + x – 4
Now
you
try .
.
.
Write a quadratic function whose graph has the given characteristics.
1. vertex: (4, –5)
y = (x – 4)2 – 5
passes through: (2, –1)
2. (–1, 5), (0, –1), (2, 11)
3. x-intercepts: –2, 5
passes through: (6, 2)
y = 4x2 – 2x – 1
y=
1 (x + 2)(x – 5)
4
Homework 4.10
pg 312: 3-39