College Algebra

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Previous Lessons
Open Chapter Ch. 1: Graphs, Functions, and Models
Lesson #1 Graphing
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Lesson #2 Graphing Functions
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Lesson #3 Slope of Linear Functions
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Lesson #4 Linear Equations and Modeling
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Lesson #5 Linear Equations, Functions, Zeros and Applications
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Lesson #6 Linear Inequalities
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Open Chapter Ch. 2: More on Functions
Lesson #7 Analyzing Functions
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Exam Exam 1
Lesson #8 Algebra for Functions
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Lesson #9 The Composition of Functions
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Lesson #10 Symmetry
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Lesson #11 Transformations
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Lesson #12 Equations of Variation
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Open Chapter Ch. 3: Quadratic Functions and Equations; Inequalities
Lesson #13 Complex Numbers
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Lesson #14 Quadratic Equations, Functions, Zeros and Models
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Lesson #15 Analyzing Graphs of Quadratic Functions
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Lesson #16 Rational and Radical Equations
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Exam Midterm Exam
Lesson #17 Absolute Value Equations and Inequalities
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Open Chapter Ch. 4: Polynomial Functions and Rational Functions
Lesson #18 Polynomial Functions and Models
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Lesson #19 Graphing Polynomial Functions
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Lesson #20 Polynomial Division
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Lesson #21 The Zeros of Polynomial Functions
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Lesson #22 Rational Functions
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Lesson #23 Polynomial and Rational Inequalities
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Open Chapter Ch. 5: Exponential Functions and Logarithmic Functions
Lesson #24 Inverse Functions
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Lesson #25 Exponential Functions and Graphs
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Lesson #26 Logarithmic Functions and Graphs
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Exam Exam 3
Lesson #27 Properties of Logarithmic Functions
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Lesson #28 Solving Exponential Equations and Logarithmic Equations
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Lesson #29 Applications and Models: Growth and Decay; Compound Interest
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Open Chapter Ch. 6: Systems of Equations
Lesson #30 Systems of Equations in Two Variables
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Lesson #31 Systems of Inequalities and Linear Programming
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Lesson #32 Nonlinear Systems of Equations and Inequalities
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Lesson #33 Sequences and Series
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Exam Final Exam

Assignments:

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Lesson Objectives:

- Find the vertex, axis of symmetry, and the maximum or minimum of a quadratic function.
- Graph quadratic functions.
- Solve an applied problem involving the maximum or minimum of a quadratic function.


The graph of a quadratic function is called a parabola.

The graph of every parabola can be obtained from the squaring function, `f(x) = x^2`, using transformations.

`f(x) = a(x-h)^2+k` is a parabola. The parabola opens up if a is positive, and the parabola opens down if a is negative.

The vertex is at the point (h, k) and the axis of symmetry is at `x = h`.

The minimum value is k when the parabola opens up, and the maximum value is k when the parabola opens down.



Find the vertex, axis of symmetry, and the maximum or minimum value and then graph the quadratic function.

We're given `f(x) = x^2-8x+13`. We need to get this in the form `f(x) = a(x-h)^2+k`. And we're going to do this by completing the square. So `f(x) = x^2-8x+(-8/2)^2-(-8/2)^2+13`. This is `x^2-8x+16-16+13`, so we have `(x-4)^2-16+13`. This is `(x-4)^2-3`. So `a = 1`, `h = 4`, and `k = -3`.

So that means that our vertex is at the point (4, -3), and our axis of symmetry is at `x = 4`. And since a is positive, our parabola opens up, and we have a minimum at k, which is -3.

So if we graph this: here's our vertex, (4, -3), then our axis of symmetry, `x = 4`, and our parabola opens up, because a is positive.



Find the vertex, axis of symmetry, and the maximum or minimum value, and then graph the quadratic function.

We're given `f(x) = (1/2)x^2+4x+13`. Start by factoring `1/2` out of our first two terms. This gives us `1/2(x^2+8x)+13`.

Now let's complete the square inside of our parentheses. That means that we're adding `(8/2)^2` and subtracting `(8/2)^2`. `(8/2)^2 = 16`. So we have `1/2(x+4)^2+1/2(-16)+13`. That's `-8+13`, which is 5. So we have `1/2(x+4)^2+5`. So compare this to `a(x-h)^2+k`. We have `a = +1/2`, `h = -4`, and `k = 5`. That means that we have a vertex at the point (-4, 5), and then we have the axis of symmetry at `x = -4`, and then we have a parabola that opens up, because a is positive. The minimum is 5.

So this is the graph of our parabola.



Find the vertex, axis of symmetry, and the maximum or minimum value, and then graph the quadratic function.

`f(x) = -x^2-6x+9`. So we need to start by factoring out the negative value from our first two terms. This gives us `f(x) = -1(x^2+6x)+9`. And then complete the square, by adding and subtracting `(6/2)^2` on the inside of our parentheses.

So we get `-1(x^2+6x+9-9)+9`. So we get `-1(x+3)^2+(-1)(-9)+9`. So we have 18 on the outside of our parentheses. `-1(x+3)^2+18`. Now comparing this to `a(x-h)^2+k`, we know that `a = -1`, `h = -3`, and `k = 18`.

So that our vertex is at the point (-3, 18), and our axis of symmetry is `x = -3`. And since a is negative, it opens down. That means that we have a maximum value, and it's 18.

So let's go ahead and graph this. The vertex is at (-3, 18), and our axis of symmetry is `x = -3`, and the parabola opens down.



Given `f(x) = ax^2+bx+c`, the vertex is at `(-b/(2a), f(-b/(2a)))`.

This means that you solve for the x-value of your vertex first, and then plug back into your function to find your y-value.



Find the vertex, the range, the maximum or minimum value, and the increasing and decreasing intervals.

We're given `f(x) = x^2-6x+9`. So the vertex; it's `(-b/(2a), f(-b/(2a)))`.

`-b/(2a) = -(-6)/(2(1))`, which is `6/2`, which is 3.

And then `f(3) = 3^2-6(3)+9` or `9-18+9` or `9-9`, which is zero.

So the vertex is at the point (3, 0). a is positive 1, so our parabola opens up. This means that our range is `[0, oo)`. And notice that we have a bracketed zero, because it includes the point (3, 0). The minimum is 0. It's increasing on the interval `(3, oo)`, and it's decreasing on the interval `(-oo, 3)`.



A ball that is thrown straight up from a height of 5.5 feet and an initial velocity of 15 feet per second is s(t) feet from the ground after t seconds.

`s(t) = 16t^2+15t+5.5`. What is the maximum height that the ball travels before falling back to the earth, and when does the ball reach its maximum height?

So the maximum height occurs at the vertex. The vertex is at the point, `t = -b/(2a)`, which is `-15/(2(16))`; this is equal to about 0.469 seconds. And then `s(0.469)` is the maximum height that the ball travels. It's equal to `16(0.469)^2+15(0.469)+5.5`. This is 16.05 feet.

The ball reaches a maximum height of 16.05 feet after 0.469 seconds.