Download Chapter 10-1 Identify Quadratics and their graphs A parabola is the

Chapter 10-1 Identify Quadratics and their graphs
A parabola is the graph of a quadratic function. A quadratic function is a function that can be written
in the form, f(x) = ax2 + bx + c, a ≠ 0 or y = ax2 + bx + c, a ≠ 0 This form is the standard form of a
quadratic function.
When a quadratic function is in the standard form y = ax2 + bx + c, a ≠ 0, the value of the leading
coefficient, a, determines the direction that the parabola opens.
 When a > 0 the parabola opens upward.
 When a < 0 the parabola opens downward.
Positive Quadratic y = x2
Negative Quadratic y = -x2
The point at which the parabola changes directions is called the vertex of the parabola.
 If the parabola opens upward the vertex is the lowest point or minimum value of the function.
 If the parabola opens downward the vertex is the highest point or maximum value of the
function.
The domain of a quadratic function f(x) = ax2 + bx + c, a ≠ 0, is the set of all real numbers.
 If the parabola opens upward the range of the function will be all values greater than or equal to
the minimum value.
 If the parabola opens downward the range of the function will be all values less than or equal to
the maximum value.
You can fold the graph of any quadratic function y = ax2 + bx + c, a ≠ 0 at the vertical line through its
vertex, and the two halves match exactly. This fold line is called the axis of symmetry. The vertex of a
parabola is the only point on the parabola that is on the axis of symmetry.
The x-intercepts of the graph are the x-values for the points where the graph intersects the x-axis. A
parabola may have one, two, or no x-intercepts.
Chapter 10-2 Graph Quadratic Functions: Parabola
You can graph a quadratic function by making a function table. It is helpful to always include the
vertex as one of the points in the table when graphing a quadratic function.
For the graph of a quadratic function f(x) = ax2 + bx + c, or y = ax2 + bx + c, where a, b and c are real
numbers and a ≠ 0.
 x = -(b/2a) is the equation of its axis of symmetry
 The x-coordinate of its vertex is -(b/2a), to find the y-value of the vertex substitute the x-value
into the original equation.
To graph a quadratic function:
 Find the equation of the axis of symmetry
 Find the coordinates of the vertex
 Make a function table. Select three x values greater than the x-coordinate of the vertex and
three x values less than the x-coordinate of the vertex. The more points in your table the more
accurate your graph.

Example:
Graph y = x2 – 8x + 12
1. Find the axis of symmetry x = -(-8/2*1)) = 8/2 = 4
2. Find the coordinates of the vertex. The x coordinate = 4, the y-coordinate is:
y = (4)2 – 8(4) +12 = 16 – 32 + 12 = -4
Vertex is (4,-4)
3. Make a function table
4. Graph the ordered pairs in the table on the coordinate plane. Draw a smooth curve through
them.
x
y = x2 – 8x + 12
(x,y)
1
(1)2 – 8(1) +12 = 5
(1,5)
2
(2)2 – 8(2) +12 = 0
(2,0)
3
4
5
2
(3,-3)
2
(4,-4)
2
(5,-3)
2
(3) – 8(3) + 12 = -3
(4) – 8(4) +12 = -4
(5) – 8(5) + 12 = -3
6
(6) – 8(6) + 12 = 0
(6,0)
7
(7)2 – 8(7) + 12 = 5
(7,5)
Chapter 10.3 – Solving Quadratic Equations by Factoring
The equation x2 +x – 20 = 0 is a quadratic equation in one variable that is in standard form.
The equation ax2 + bx + c = 0 where a, b and c are real numbers and a ≠ 0 is a quadratic equation that is
written in standard form. The values of the variable in an equation that make the equation true are
called solutions of the equation.
We can solve a quadratic equation by factoring. To do this we must apply the Zero Product Property
which states that if ab = 0 then either a = 0 or b = 0, where a and b are real numbers. This
property enables you to conclude that if a product is equal to 0, then at least one of its factors must be
equal to 0.
In our example x2 +x – 20 = 0 if we factor the quadratic we get (x+5)(x-4) = 0. The Zero Product
Property states that one of the factors must equal 0. So either (x+5) = 0 or (x-4) = 0. We can solve
these equations to determine that either x = -5 or x = 4. So the solution set for x2 +x – 20 = 0 is {-5,4}.
The solutions of the quadratic equation, ax2 + bx + c = 0, are the x-intercepts of the related quadratic
function, f(x) = ax2 + bx + c. The solutions of the equation are called the roots of the equation or zeros
of the function.
Sometimes there is only one solution to a quadratic equation. In this case the x-intercept is also
the vertex of the quadratic function.
Not every quadratic equation can be factored over integers. Sometimes the roots of a quadratic
equation are irrational numbers, or the roots are not real numbers. The x-intercepts of quadratics that
cannot be factored can be approximated through graphing.
If a quadratic does not have x-intercepts then it does not have real roots.
Some equations containing radicals result in quadratic equations. Sometimes the quadratic equations
will have extra roots, called extraneous roots, so not all the roots of the quadratic equations are
solutions to the original equations.
Solutions to quadratic equations are given in set notation.
One important issue should be mentioned at this point: Just as with linear equations, the solutions to quadratic
equations may be verified by plugging them back into the original equation, and making sure that they work, that
they result in a true statement. For the above example, we would do the following:
Checking x = 3 in (x – 3)(x – 4) = 0:
([3] – 3)([3] – 4) ?=? 0
(3 – 3)(3 – 4) ?=? 0
(0)(–1) ?=? 0
0 = 0
Checking x = 4 in (x – 3)(x – 4) = 0:
([4] – 3)([4] – 4) ?=? 0
(4 – 3)(4 – 4) ?=? 0
(1)(0) ?=? 0
0 = 0
Chapter 10.4 – Solve Verbal Problems Involving the Quadratic Equations
Mrs. Baca's art class is painting a mural on the front of the school. The mural is 3 meters wider than it
is high and has an area of 10m2. What are the dimensions of the mural.
To solve this problem you may want to begin by drawing a sketch of the mural.
We can let x = the height of the of the mural, so the width will be x+3. The area is 10m2. So we can
set up the equation 10m2= x(x+3)
If we multiply this we get 10 = x2 + 3x. We then want to write the equation in standard form by
subtracting 10 from each side of the equation.
x2 + 3x – 10 = 0 Now we want to factor the quadratic
(x+5)(x-2) = 0 Now we want to apply the Zero Product Property
x+5 = 0 or x-2 = 0
x = -5 or x = 2
You know that since x represents the height of the mural so x = -5 can be eliminated as a solution for x
since you cannot have a negative height. The height of the mural will be 2 meters and the width will be
5 meters.
You can write a quadratic equation, given its roots, by working backward.
 From the roots write the binomial factors
 Multiply the factors to write the equation in standard form.
Example: Write an equation with given roots {-3,5}
If the roots are -3 and 5 then the factors are (x+3) and (x-5)
(x+3)(x-5) = 0
x2+3x-5x-15 = 0
x2 – 2x – 15 = 0 is an equation with roots {-3,5}
Chapter 10.5 – Solve Quadratic Equations by Completing the Square
To solve a quadratic equation by completing the square you must make the quadratic expression on one
side of the equation into a perfect square. After completing the square you can solve the equation by
taking the square root of each side.
Completing the square is often a good method to use for solving a quadratic equation when the
equation is not factorable using intergers.
Completing the Square to solve ax2 + bx + c = 0 where a, b and c are real numbers and a ≠ 0:
 Write the equation so that the constant term, c, is isolated on the right side. Divide each side of
the equation by a; x2 +(b/a)x = c/a
 Find the square of one half the coefficient of x. Add that number to each side of the equation.
 Factor the left side of the equation. The results should have the form (x + r)2 where r is a
constant.
 Take the square root of each side. Then solve for x, and simplify the solutions.
Example: Solve 3x2 – 7x +2 = 0
3x2 – 7x = -2
rewrite equation with constant isolated on right side
of equation
x2 – 7/3x = -2/3
divide both sides of the equation by a
½ of 7/3 = 7/6
Find ½ the value of the x term
x2 – 7/3x + 49/36 = -2/3 + 49/36
add the squared value of ½ x to both sides of the
equation.
(x-7/6)2 = -24/36 + 49/36
make like denominators and add the values on the
side of equation
(x-7/6)2 = 25/36
rewrite the trinomial as the square of a binomial
√(x-7/6)2 = √25/36
take the square root of both sides of the equation
x-7/6 = 5/6
or
x-7/6 + 7/6 = 5/6 +7/6
x = 12/6 = 2
x – 7/6 = – 5/6
write and solve two equations.
x – 7/6 + 7/6 = -5/6 + 7/6
x = 2/6 = 1/3
So, x = 2 or x = 1/3
It is important to remember that every positive real number has two square roots – one positive and one
negative.
Chapter 10.6 – The Quadratic Formula and the Discriminant
The Quadratic Formula is a formula that can be used to solve any quadratic equation. It is derived by
solving the standard form of a quadratic equation, ax2 + bx + c = 0, for x by completing the square.
The Quadratic Formula is as follows:
If ax2 + bx + c = 0, where a ≠ 0, then x = -b +/- √b2- 4ac
2a
The expression b2- 4ac is called the discriminant of the quadratic equation, and it provides important
information about the roots of the equation.
The symbol +/- indicates that the discriminant will be both added and subtracted, so the equation will
have two roots.

If b2- 4ac = 0 then the only root of the equation is -b/2a, which is a rational number.

If b2- 4ac > 0 then – b +/- √positive number. If b2- 4ac is a perfect square, both roots will be
2a
rational. If it is not, the two roots will be irrational.

If b2- 4ac < 0 then x = -b +/- √negative number. The equation will have no real roots because
2a
the square root of a negative number is not a real number.
Chapter 10.7 – Solve Quadratic Equations with the Quadratic Formula
The equation 3x2 + 40x – 275 can be solved by factoring or completing the square, however a more
practical method would be to use the Quadratic Formula.
Remember that the Quadratic Formula is:
If ax2 + bx + c = 0, where a ≠ 0, then x = -b +/- √b2- 4ac
2a
In our example: a = 3, b = 40 and c = -275
x = -40 +/- √(40)2 – 4(3)(-275)
2(3)
x = -40 +/- √1600 – (-3300)
6
x = -40 +/- √4900
6
x = -40 +/- 70
6
x = -40 + 60 = 30
6
6
x = 5 or x = -55/3
or x = -40 – 70 = -110
6
6
Chapter 10.8 – Solve Linear-Quadratic Systems
One way to solve a system of linear-quadratic equations is by substitution. Solve the linear equation
for y and substitute that value into the quadratic equation.
Let’s try it:
y = x2 + 5x + 18
y = -2x + 18
y = -2x + 18 is already solved for y so all we have to do is substitute the value for y in the linear
equation into the quadratic equation.
-2x + 18 = x2 + 5x + 18
+2x
+2x
18 = x2 + 7x +18
-18
- 18
0 = x2 + 7x
addition property of equality
subtraction property of equality
0 = x(x+7) distributive property
x = 0 or x+7 = 0 zero product property
x = 0 or x = -7
Now to find y we have to substitute the values for x into either equation.
If x = 0 then y = -2(0) + 18, y = 18
If x = -7, then y = -2(-7) + 18, y = 14 + 18, y = 32
(0,18) is a solution and (7,-32) is a solution
Systems of linear-quadratic equations can have two solutions (two points of intersection), one solution
(the parabola and line intersect at one point), or no solutions (no intersection).
When you solve a linear-quadratic system you must check your solution in both equations in the
system. What you are doing when you solve a system is finding the point or points where the two
graphs intersect. That point or points when substituted into the original equations will make them both
true. If the point does not make both equations true it is not a solution to the system.
When solving a linear-quadratic system of equations:
Solve one equation for one variable in terms of the other
Substitute into the equation not used in step 1
Solve the equation
Substitute to find the corresponding values of the other variable
Check the solutions