The Central Idea
A quadratic function has the form
\[ \boxed{f(x)=ax^2+bx+c}, \]where
\[ a\neq0. \]The condition \(a\neq0\) is essential. If \(a=0\), the \(x^2\) term disappears and the function is no longer quadratic.
The graph of every quadratic function is a parabola.
Quadratic Functions Extend the Power Function \(x^2\)
The simplest quadratic power function is
\[ y=x^2. \]A general quadratic
\[ y=ax^2+bx+c \]adds scaling, reflection, horizontal placement, and vertical placement to that basic parabolic shape.
The quadratic family therefore grows directly from the power-function ideas developed in the previous lesson.
Quadratic Functions and Quadratic Equations
A quadratic function assigns outputs:
\[ f(x)=ax^2+bx+c. \]A quadratic equation asks for values of \(x\) that make two expressions equal.
A common form is
\[ ax^2+bx+c=0. \]Solving this equation means finding the zeros of the quadratic function.
The Parabola
A parabola is the graph of a quadratic function.
For example,
\[ y=x^2 \]produces the familiar upward-opening parabola.
If the leading coefficient is negative, as in
\[ y=-x^2, \]the graph is reflected across the \(x\)-axis and opens downward.
The Role of the Leading Coefficient
In
\[ f(x)=ax^2+bx+c, \]the coefficient \(a\) controls the parabola's orientation and vertical scale.
- If \(a>0\), the parabola opens upward.
- If \(a<0\), the parabola opens downward.
- If \(|a|>1\), the graph is vertically stretched.
- If \(0<|a|<1\), the graph is vertically compressed.
The \(y\)-Intercept
In standard form,
\[ f(x)=ax^2+bx+c, \]evaluate at \(x=0\):
\[ f(0)=c. \]Therefore, the \(y\)-intercept is
\[ \boxed{(0,c)}. \]The constant term can be read directly from the graph as the vertical-axis intercept.
Zeros and \(x\)-Intercepts
A zero of the quadratic function satisfies
\[ f(x)=0. \]Therefore, solving
\[ ax^2+bx+c=0 \]finds the \(x\)-coordinates where the parabola intersects the \(x\)-axis.
Real zeros correspond to real \(x\)-intercepts.
Factored Form
If a quadratic factors as
\[ f(x)=a(x-r_1)(x-r_2), \]then the zeros are visible immediately:
\[ \boxed{x=r_1} \]and
\[ \boxed{x=r_2}. \]Factored form is especially useful when studying roots and \(x\)-intercepts.
Example: Read Zeros from Factored Form
Consider
\[ f(x)=(x-3)(x+2). \]Set each factor equal to zero:
\[ x-3=0 \]or
\[ x+2=0. \]Therefore,
\[ \boxed{x=3,\ -2}. \]The graph has \(x\)-intercepts
\[ \boxed{(3,0)} \]and
\[ \boxed{(-2,0)}. \]Standard Form
The standard form
\[ \boxed{f(x)=ax^2+bx+c} \]makes the coefficients and \(y\)-intercept easy to identify.
It is especially useful for:
- reading the leading coefficient;
- reading the constant term;
- using the axis-of-symmetry formula;
- connecting to the quadratic formula later.
Vertex Form
A quadratic may also be written as
\[ \boxed{f(x)=a(x-h)^2+k}. \]This is called vertex form.
The vertex is
\[ \boxed{(h,k)}. \]The axis of symmetry is
\[ \boxed{x=h}. \]The Vertex
The vertex is the turning point of a parabola.
If \(a>0\), the vertex is the minimum point.
If \(a<0\), the vertex is the maximum point.
Every parabola is symmetric about the vertical line passing through its vertex.
Axis of Symmetry
For
\[ f(x)=ax^2+bx+c, \]the axis of symmetry is
\[ \boxed{x=-\frac{b}{2a}}. \]This formula gives the \(x\)-coordinate of the vertex.
Finding the Vertex from Standard Form
Consider
\[ f(x)=x^2-6x+5. \]Here,
\[ a=1,\qquad b=-6. \]The axis of symmetry is
\[ x=-\frac{-6}{2(1)}=3. \]Evaluate:
\[ f(3)=9-18+5=-4. \]Therefore, the vertex is
\[ \boxed{(3,-4)}. \]Why the Axis Lies Halfway Between Two Real Zeros
Suppose a quadratic has real zeros
\[ r_1 \]and
\[ r_2. \]Because the parabola is symmetric, the axis of symmetry lies halfway between the two intercepts.
Therefore,
\[ \boxed{ x=\frac{r_1+r_2}{2} }. \]Example: Find the Axis from the Zeros
Suppose the zeros are
\[ -2 \]and
\[ 6. \]The axis lies halfway between them:
\[ x=\frac{-2+6}{2}=2. \]Therefore,
\[ \boxed{x=2}. \]The Three Main Forms
| Form | Expression | Information Revealed |
|---|---|---|
| Standard | \(ax^2+bx+c\) | Coefficients, \(y\)-intercept, axis formula |
| Factored | \(a(x-r_1)(x-r_2)\) | Zeros and \(x\)-intercepts |
| Vertex | \(a(x-h)^2+k\) | Vertex and axis of symmetry |
These are not different quadratic functions. They are different representations of the same function.
Moving from Factored Form to Standard Form
Consider
\[ f(x)=2(x-1)(x+3). \]Multiply:
\[ (x-1)(x+3)=x^2+2x-3. \]Then
\[ f(x)=2x^2+4x-6. \]Therefore,
\[ \boxed{ 2(x-1)(x+3)=2x^2+4x-6 }. \]Moving from Vertex Form to Standard Form
Consider
\[ f(x)=3(x-2)^2-5. \]Expand the square:
\[ (x-2)^2=x^2-4x+4. \]Multiply:
\[ 3x^2-12x+12. \]Subtract \(5\):
\[ \boxed{f(x)=3x^2-12x+7}. \]Solving a Quadratic Equation by Factoring
Solve
\[ x^2-5x+6=0. \]Factor:
\[ (x-2)(x-3)=0. \]By the zero-product principle,
\[ x-2=0 \]or
\[ x-3=0. \]Therefore,
\[ \boxed{x=2,\ 3}. \]Why the Equation Must Be Set Equal to Zero
The zero-product principle applies when a product equals zero.
Consider
\[ x^2+3x=10. \]Move all terms to one side:
\[ x^2+3x-10=0. \]Then factor:
\[ (x+5)(x-2)=0. \]Therefore,
\[ \boxed{x=-5,\ 2}. \]Solving by Square Roots
Some quadratic equations can be solved directly from
\[ x^2=k. \]For example,
\[ x^2=49. \]Taking square roots gives
\[ \boxed{x=\pm7}. \]Both signs are necessary because
\[ 7^2=(-7)^2=49. \]Example: Isolate the Square First
Solve
\[ 3x^2-12=0. \]Add \(12\):
\[ 3x^2=12. \]Divide by \(3\):
\[ x^2=4. \]Therefore,
\[ \boxed{x=\pm2}. \]Equations of the Form \((x-h)^2=k\)
A shifted square can also be solved by square roots.
Suppose
\[ (x-3)^2=16. \]Then
\[ x-3=\pm4. \]Therefore,
\[ x=3\pm4. \]So
\[ \boxed{x=7,\ -1}. \]Quadratic Equations Can Have Different Numbers of Real Solutions
A quadratic equation may have:
- two distinct real solutions;
- one repeated real solution;
- no real solutions.
The graph gives a visual interpretation of these cases.
Two Real Solutions
If a parabola crosses the \(x\)-axis at two distinct points, then
\[ f(x)=0 \]has two distinct real solutions.
For example,
\[ x^2-5x+6=0 \]has solutions
\[ 2,\ 3. \]One Repeated Real Solution
If the vertex lies exactly on the \(x\)-axis, the parabola touches the axis at one point.
For example,
\[ x^2-6x+9=0 \]factors as
\[ (x-3)^2=0. \]The only real solution is
\[ \boxed{x=3}. \]It is a repeated root.
No Real Solutions
Consider
\[ x^2+1=0. \]This would require
\[ x^2=-1. \]No real number has square \(-1\).
Therefore, the equation has no real solutions.
Later we will extend the number system to complex roots.
The Vertex and Maximum or Minimum Value
The vertex gives the extreme value of a quadratic function.
If \(a>0\), the vertex gives the minimum value.
If \(a<0\), the vertex gives the maximum value.
For
\[ f(x)=2(x-3)^2-5, \]the vertex is
\[ (3,-5). \]Since \(a=2>0\), the minimum value is
\[ \boxed{-5}. \]Range from Vertex Form
Consider
\[ f(x)=a(x-h)^2+k. \]If \(a>0\), then
\[ f(x)\ge k. \]Therefore,
\[ \boxed{\text{Range}=[k,\infty)}. \]If \(a<0\), then
\[ f(x)\le k, \]so
\[ \boxed{\text{Range}=(-\infty,k]}. \]Domain of a Quadratic Function
Every polynomial quadratic function is defined for all real inputs.
\[ \boxed{\text{Domain}=(-\infty,\infty)}. \]End Behavior of Quadratic Functions
Every quadratic has even degree.
Therefore, both ends of the graph move in the same direction.
If \(a>0\),
\[ f(x)\to\infty \]at both ends.
If \(a<0\),
\[ f(x)\to-\infty \]at both ends.
A Complete Example from Standard Form
Analyze
\[ f(x)=x^2-4x-5. \]The leading coefficient is positive, so the parabola opens upward.
The \(y\)-intercept is
\[ \boxed{(0,-5)}. \]Factor:
\[ x^2-4x-5=(x-5)(x+1). \]Therefore, the zeros are
\[ \boxed{x=5,\ -1}. \]Continue the Complete Example
The axis of symmetry lies halfway between the zeros:
\[ x=\frac{5+(-1)}{2}=2. \]Evaluate:
\[ f(2)=4-8-5=-9. \]Therefore, the vertex is
\[ \boxed{(2,-9)}. \]Since the parabola opens upward, the minimum value is
\[ \boxed{-9}. \]Quadratic Modeling
Quadratic functions appear in many applied settings.
They can model:
- projectile height under constant gravitational acceleration;
- areas formed from changing dimensions;
- revenue or profit under simplified pricing assumptions;
- optimization problems involving maximum or minimum values;
- geometric relationships containing squared lengths.
Example: Area Model
Suppose a rectangle has dimensions
\[ x \]and
\[ x+5. \]Its area is
\[ A(x)=x(x+5). \]Therefore,
\[ \boxed{A(x)=x^2+5x}. \]The area is a quadratic function of \(x\).
Example: Projectile Height
A simplified vertical-motion model might have the form
\[ h(t)=-16t^2+64t+5, \]where \(h\) is height in feet and \(t\) is time in seconds.
The negative leading coefficient means the graph opens downward.
Therefore, the function has a maximum height at its vertex.
Quadratic models naturally encode both growth and decline around a turning point.
Function Language and Equation Language
It is useful to distinguish these questions:
“What is the value of the function at \(x=3\)?”
versus
“For which values of \(x\) does the function equal zero?”
The first asks for evaluation.
The second asks for solutions of a quadratic equation.
Common Error: Treating \(ax^2+bx+c\) as Linear
The presence of the \(x^2\) term changes the function family.
A quadratic graph is not a straight line.
Its rate of change varies with \(x\), and the graph bends into a parabola.
Common Error: Reading the Vertex from Standard Form
In
\[ ax^2+bx+c, \]the vertex is not generally
\[ (b,c). \]The \(x\)-coordinate of the vertex must be found from
\[ \boxed{x=-\frac{b}{2a}}, \]and then substituted back into the function.
Common Error: Forgetting Both Square-Root Solutions
If
\[ x^2=25, \]then
\[ \boxed{x=\pm5}. \]Writing only \(x=5\) loses the negative solution.
Common Error: Applying the Zero-Product Principle Too Early
The equation
\[ (x-2)(x+3)=10 \]does not imply
\[ x-2=0 \]or
\[ x+3=0. \]The zero-product principle applies only when the product equals zero.
A Reliable Procedure for Graphing a Quadratic
- Identify the sign of \(a\) to determine opening direction.
- Find the axis of symmetry.
- Find the vertex.
- Find the \(y\)-intercept.
- Find real zeros if possible.
- Use symmetry to locate matching points.
- Sketch the parabola through the known points.
A Reliable Procedure for Solving a Quadratic by Factoring
- Move all terms to one side so the equation equals zero.
- Factor out any GCF.
- Factor the quadratic completely.
- Set each factor equal to zero.
- Solve the resulting linear equations.
- Substitute back if verification is useful.
A Reliable Procedure for Solving by Square Roots
- Isolate the squared expression.
- Take both positive and negative square roots.
- Solve the resulting equations.
- Check the solutions when appropriate.
Practice 1: Identify the Structure
For
\[ f(x)=-2x^2+5x+3, \]identify \(a\), \(b\), \(c\), the opening direction, and the \(y\)-intercept.
Practice 2: Solve by Factoring
Solve:
\[ x^2-7x+12=0. \]Practice 3: Solve After Rearranging
Solve:
\[ x^2+2x=15. \]Practice 4: Solve by Square Roots
Solve:
\[ 2x^2=50. \]Practice 5: Shifted Square
Solve:
\[ (x+4)^2=9. \]Practice 6: Find the Vertex
Find the vertex of
\[ f(x)=x^2-8x+7. \]Practice 7: Read Factored Form
For
\[ f(x)=2(x-5)(x+1), \]identify the zeros and the axis of symmetry.
Practice 8: Read Vertex Form
For
\[ f(x)=-3(x-2)^2+7, \]identify the vertex, axis of symmetry, opening direction, and maximum or minimum value.
Practice 9: Number of Real Solutions
Without using the quadratic formula, determine how many real solutions each equation has:
- \((x-2)(x+5)=0\)
- \((x-3)^2=0\)
- \(x^2+4=0\)
Practice 10: Complete Function Analysis
Analyze
\[ f(x)=x^2-2x-8. \]Find:
- opening direction;
- \(y\)-intercept;
- zeros;
- axis of symmetry;
- vertex.
Practice Answers
Practice 1
For
\[ -2x^2+5x+3, \]- \(a=\boxed{-2}\);
- \(b=\boxed{5}\);
- \(c=\boxed{3}\);
- the parabola opens downward;
- the \(y\)-intercept is \(\boxed{(0,3)}\).
Practice 2
\[ x^2-7x+12=(x-3)(x-4). \]Therefore,
\[ \boxed{x=3,\ 4}. \]Practice 3
Rearrange:
\[ x^2+2x-15=0. \]Factor:
\[ (x+5)(x-3)=0. \]Therefore,
\[ \boxed{x=-5,\ 3}. \]Practice 4
\[ 2x^2=50 \] \[ x^2=25 \] \[ \boxed{x=\pm5}. \]Practice 5
\[ x+4=\pm3. \]Therefore,
\[ \boxed{x=-1,\ -7}. \]Practice 6
The axis is
\[ x=-\frac{-8}{2}=4. \]Evaluate:
\[ f(4)=16-32+7=-9. \]Therefore, the vertex is
\[ \boxed{(4,-9)}. \]Practice 7
The zeros are
\[ \boxed{5,\ -1}. \]The axis lies halfway between them:
\[ x=\frac{5+(-1)}{2}=2. \]Therefore,
\[ \boxed{x=2}. \]Practice 8
Vertex:
\[ \boxed{(2,7)}. \]Axis:
\[ \boxed{x=2}. \]Since \(a=-3<0\), the parabola opens downward and has maximum value
\[ \boxed{7}. \]Practice 9
- Two distinct real solutions.
- One repeated real solution.
- No real solutions.
Practice 10
For
\[ f(x)=x^2-2x-8, \]- opens upward;
- \(y\)-intercept: \(\boxed{(0,-8)}\);
- factorization: \((x-4)(x+2)\);
- zeros: \(\boxed{4,\ -2}\);
- axis: \(x=\frac{4+(-2)}2=\boxed{1}\);
- vertex: \(f(1)=1-2-8=-9\), so \(\boxed{(1,-9)}\).
What to Remember
A quadratic function has the form
\[ \boxed{f(x)=ax^2+bx+c,\qquad a\neq0}. \]Its graph is a parabola.
The three principal forms are:
\[ \boxed{ax^2+bx+c} \]for standard form,
\[ \boxed{a(x-r_1)(x-r_2)} \]for factored form, and
\[ \boxed{a(x-h)^2+k} \]for vertex form.
The axis of symmetry is
\[ \boxed{x=-\frac{b}{2a}}. \]Quadratic equations can often be solved by factoring or by isolating a square and taking square roots.
Real solutions are the \(x\)-coordinates where the parabola meets the \(x\)-axis.
Where We Go Next
Not every quadratic factors conveniently, and standard form does not make the vertex immediately visible. A systematic transformation can solve both problems.
The next lesson, Completing the Square and the Quadratic Formula, develops completing the square, derives vertex form from standard form, and uses that same process to derive the quadratic formula.