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Quadratic Functions Section 3.1 SLU Mat 151 Spring 2007

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**1. **Quadratic Functions Section 3.1 SLU Mat 151 Spring 2007

**3. **Properties of Quadratic Functions 1) If the coefficient a of the quadratic term is positive, the parabola opens upward; otherwise, the parabola opens downward.
2)The vertex (or turning point) is the minimum or maximum point.
3) The domain of any quadratic is the set of all real numbers.
4) The range will depend on the leading coefficient a.
5) Each have an axis of symmetry.

**4. **The Standard Form of a Quadratic Function The quadratic function
f (x) = a (x - h)2 + k, a ? 0
is in standard form.
The vertex is the point (h, k).
The parabola is symmetric to the line x = h.

**5. **Graphing Parabolas With Equations in Standard Form To graph f (x) = a (x - h)2 + k:
Determine whether the parabola opens upward or downward. If a > 0, it opens upward. If a < 0, it opens downward.
Determine the vertex of the parabola. The vertex is (h, k).
Find any x-intercepts by replacing f (x) with 0. Solve the resulting quadratic equation for x.
Find the y-intercept by replacing x with zero.
Plot the intercepts and vertex. Connect these points with a smooth curve that is shaped like a cup.

**6. **Example Graph the quadratic function
f (x) = -2(x - 3)2 + 8
Solution: Let’s follow our steps to do this!

**7. **Example cont. Step 1: a = -2 which means that the graph open down
Step 2: Vertex(3, 8)
Step 3: To find the x-intercept we set f(x) = 0.
and solve for x:
-2(x - 3)2 + 8 = 0

**8. **Example Cont. -2(x - 3)2 + 8 = 0
-2(x - 3)2 = -8
(x - 3)2 = 4
x – 3 = 2 or x - 3 = -2
x = 5 or x = 1
The x-intercepts are 5 and 1.
Step 4: The y-intercept is f(0) = -2(0 - 3)2+8=10
Step 5: Graph ?!

**9. **The quadratic functions in General form: f (x) = ax 2 + bx + c
The vertex is at

**10. **Example Solution:
Step 1 Determine how the parabola opens. Note that a, the coefficient of x 2, is -1. Thus, a < 0; this negative value tells us that the parabola opens downward.
Step 2 Find the vertex. We know the x-coordinate of the vertex is x = -b/2a.
x = -b/2a = -6/2(-1)=3.
The x-coordinate of the vertex is 3. We substitute 3 for x in the equation of the function to find the y-coordinate:
y = f(3) = –(3)2 + 6(3) – 2 = – 9 + 18 – 2 = 7, the parabola has its vertex at (3,7).

**11. **Example Step 3 Find the x-intercepts. Replace f (x) with 0 in f (x) = -x2 + 6x -2.
0 = -x2 + 6x - 2

**12. **Example Cont.
Step 4 The y-intercept is
f (0) = -02 + 6 • 0 - 2 = -2
Step 5 Graph the parabola.

**13. **Minimum and Maximum: Quadratic Functions Consider f(x) = ax2 + bx +c.
If a > 0, then f has a minimum that occurs at x = -b/2a.
This minimum value is f(-b/2a).
If a < 0, the f has a maximum that occurs at x = -b/2a.
This maximum value is f(-b/2a).

**14. **Cool Free Website http://www.langara.bc.ca/mathstats/resource/onWeb/precalculus/quadratics/index.htm#