Quadratic Functions

A quadratic function is a polynomial function of degree 2, represented in the general form:

\[ f(x) = ax^2 + bx + c \]

where \( a \neq 0 \), and \( a, b, c \) are real numbers.

Key Features:

Graph is a parabola
If \( a > 0 \), parabola opens upwards
If \( a < 0 \), parabola opens downwards
Vertex is the turning point
Axis of symmetry is a vertical line through the vertex

Example:

For \( f(x) = 2x^2 - 4x + 1 \):

\( a = 2 \) (opens upwards)
\( b = -4 \)
\( c = 1 \)

Graph of \( f(x) = x^2 - 4x + 3 \):

Maximum and Minimum Values

The vertex of a parabola represents either the maximum or minimum point of the quadratic function.

Vertex Formula:

The vertex \((h, k)\) can be found using:

\[ h = -\frac{b}{2a} \] \[ k = f(h) \]

Maximum Value:

When \( a < 0 \), the vertex is the highest point (maximum)

Minimum Value:

When \( a > 0 \), the vertex is the lowest point (minimum)

Example:

Find the vertex of \( f(x) = -2x^2 + 8x - 5 \):

Using the vertex formula:

\[ h = -\frac{8}{2(-2)} = 2 \] \[ k = f(2) = -2(2)^2 + 8(2) - 5 = 3 \]

Vertex is at \( (2, 3) \) and since \( a < 0 \), this is a maximum point.

Sketching Quadratic Curves

To sketch a quadratic function, follow these steps:

Determine direction: Check if \( a > 0 \) (upwards) or \( a < 0 \) (downwards)
Find the vertex: Use \( h = -\frac{b}{2a} \) and \( k = f(h) \)
Find y-intercept: Evaluate \( f(0) = c \)
Find x-intercepts (roots): Solve \( ax^2 + bx + c = 0 \)
Plot points: Vertex, intercepts, and additional points if needed
Draw parabola: Smooth curve through the points

Example:

Sketch \( f(x) = x^2 - 4x + 3 \):

\( a = 1 > 0 \) → opens upwards
Vertex: \( h = -\frac{-4}{2(1)} = 2 \), \( k = (2)^2 - 4(2) + 3 = -1 \) → \( (2, -1) \)
y-intercept: \( (0, 3) \)
x-intercepts: \( x^2 - 4x + 3 = 0 \) → \( (x-1)(x-3) = 0 \) → \( x = 1, 3 \)

(2, -1) (1, 0) (3, 0) (0, 3)

Solving Quadratic Equations

Quadratic equations in the form \( ax^2 + bx + c = 0 \) can be solved using:

1. Factorization

Express as product of two binomials:

\[ x^2 - 5x + 6 = 0 \] \[ (x-2)(x-3) = 0 \] \[ x = 2 \text{ or } 3 \]

2. Quadratic Formula

Use the formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

For \( 2x^2 + 5x - 3 = 0 \): \[ x = \frac{-5 \pm \sqrt{25 + 24}}{4} \] \[ x = \frac{-5 \pm 7}{4} \] \[ x = \frac{1}{2} \text{ or } -3 \]

3. Completing the Square

Rearrange into perfect square form:

\[ x^2 + 6x + 5 = 0 \] \[ x^2 + 6x = -5 \] \[ x^2 + 6x + 9 = 4 \] \[ (x+3)^2 = 4 \] \[ x = -3 \pm 2 \] \[ x = -1 \text{ or } -5 \]

The Discriminant

The discriminant \( D = b^2 - 4ac \) determines the nature of roots:

Discriminant	Nature of Roots
\( D > 0 \)	Two distinct real roots
\( D = 0 \)	One real root (repeated)
\( D < 0 \)	No real roots (complex)

Quadratic Inequalities

To solve quadratic inequalities like \( ax^2 + bx + c > 0 \):

Find the roots of the corresponding equation \( ax^2 + bx + c = 0 \)
Sketch the parabola based on the sign of \( a \)
Determine intervals where the inequality holds

Example:

Solve \( x^2 - 3x - 10 > 0 \):

Find roots: \( x^2 - 3x - 10 = 0 \) → \( (x-5)(x+2) = 0 \) → \( x = -2, 5 \)
Parabola opens upwards (\( a = 1 > 0 \))
Inequality holds when \( x < -2 \) or \( x > 5 \)

-2 5

Solution: \( x < -2 \)

Solution: \( x > 5 \)

Roots of Quadratic Equations

For a quadratic equation \( ax^2 + bx + c = 0 \) with roots \( \alpha \) and \( \beta \):

Sum and Product of Roots:

\[ \alpha + \beta = -\frac{b}{a} \] \[ \alpha \beta = \frac{c}{a} \]

Forming Equations:

Given roots \( \alpha \) and \( \beta \), the equation is:

\[ x^2 - (\alpha + \beta)x + \alpha \beta = 0 \]

Example:

Find a quadratic equation with roots 2 and -3:

\[ \alpha + \beta = 2 + (-3) = -1 \] \[ \alpha \beta = 2 \times (-3) = -6 \] \[ x^2 - (-1)x + (-6) = 0 \] \[ x^2 + x - 6 = 0 \]

Symmetry in Quadratic Equations

Quadratic functions exhibit symmetry about their vertex:

Axis of Symmetry:

\[ x = -\frac{b}{2a} \]

This vertical line divides the parabola into two mirror images.

Example:

For \( f(x) = x^2 - 4x + 3 \):

Axis of symmetry:

\[ x = -\frac{-4}{2(1)} = 2 \]

Points equidistant from \( x = 2 \) have the same y-value.

Axis of Symmetry

Simultaneous Equations

Solving one linear and one quadratic equation simultaneously:

Substitution Method

Solve the linear equation for one variable
Substitute into the quadratic equation
Solve the resulting quadratic
Find corresponding values of the other variable

Example:

Solve:

\[ y = x + 1 \] \[ y = x^2 - 3x + 4 \]

Substitute:

\[ x + 1 = x^2 - 3x + 4 \] \[ x^2 - 4x + 3 = 0 \] \[ (x-1)(x-3) = 0 \] \[ x = 1 \text{ or } 3 \]

Corresponding y-values:

\[ \text{When } x=1, y=2 \] \[ \text{When } x=3, y=4 \]

Solutions: \( (1,2) \) and \( (3,4) \)

Graphical Interpretation:

The solutions represent points of intersection between the line and parabola.

(1,2) (3,4)

Practice Exercise

Question 1

Find the vertex of the quadratic function \( f(x) = -x^2 + 6x - 8 \).

Using vertex formula:

\[ h = -\frac{6}{2(-1)} = 3 \] \[ k = f(3) = -(3)^2 + 6(3) - 8 = 1 \]

Vertex is at \( (3, 1) \)

Question 2

Solve the quadratic inequality \( 2x^2 - 5x - 3 \leq 0 \).

First find roots:

\[ 2x^2 - 5x - 3 = 0 \] \[ x = \frac{5 \pm \sqrt{25 + 24}}{4} = \frac{5 \pm 7}{4} \] \[ x = 3 \text{ or } -\frac{1}{2} \]

Parabola opens upwards, so \( -\frac{1}{2} \leq x \leq 3 \)

Question 3

Find the quadratic equation whose roots are \( 2 + \sqrt{3} \) and \( 2 - \sqrt{3} \).

Using sum and product of roots:

\[ \alpha + \beta = (2+\sqrt{3}) + (2-\sqrt{3}) = 4 \] \[ \alpha \beta = (2+\sqrt{3})(2-\sqrt{3}) = 4 - 3 = 1 \] \[ x^2 - 4x + 1 = 0 \]

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