Quadratic Graphs

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Notes

Key Features of Quadratic Graphs

A quadratic graph has equation **y $= ax^{2} + bx +$ c** with **a ≠ 0**.
It is a smooth curve called a **parabola**, with a vertical line of symmetry.
If **a > 0**, the graph is **u-shaped** (minimum turning point).
If **a < 0**, the graph is **n-shaped** (maximum turning point).
The **y-intercept** is at **(0, c)**.
The **x-intercepts (roots)** are solutions to **ax² $+ bx + c =$ 0**; there can be 0, 1, or 2 roots.
The **turning point (vertex)** is the minimum or maximum point.

Sketching a Quadratic Graph

Draw axes and mark the **y-intercept** (0, c).
Find and mark the **roots** by solving **ax² $+ bx + c =$ 0** (factorising, completing square, or quadratic formula).
Determine the shape: **u-shaped** if $a > 0$ , **n-shaped** if $a < 0$ .
Sketch a smooth curve through the intercepts, showing the turning point if known.
Label all intercepts and the turning point coordinates.

Finding the Turning Point by Completing the Square

Rewrite **y $= ax^{2} + bx +$ c** as **y $= a(x - p)^{2} +$ q**.
The turning point is at **(p, q)** (note sign change for p).
For **y $= (x - 3)^{2} +$ 2**, the minimum is **(3, 2)**.
For **y $= (x + 3)^{2} +$ 2**, the minimum is **(-3, 2)**.
The value of **a** does not affect the turning point coordinates but affects the shape.

Finding the Turning Point by Differentiation

Differentiate **y $= ax^{2} + bx +$ c** to get **dy/dx $= 2ax +$ b**.
Set **dy/dx = 0** and solve for **x** to find the x-coordinate of the turning point.
Substitute this x into the original equation to find the y-coordinate.
This method works for any quadratic and confirms whether it is a maximum or minimum.

Finding the Equation of a Quadratic from Its Graph

If the **vertex (p, q)** and one other point are known, use **y $= a(x - p)^{2} +$ q**.
Substitute the other point to find **a**.
If the **roots (x₁, 0) and (x₂, 0)** and one other point are known, use **y $= a(x -$ x₁)(x - x₂)**.
Substitute the other point to find **a**.
If **a = 1**, only the vertex or roots are needed.

Example: Sketching y = x² - 5x + 6

**y-intercept**: $(0, 6) (c = 6)$ .
Factorise: **y $= (x - 2)(x -$ 3)** → roots at **(2, 0)** and **(3, 0)**.
**a $= 1 >$ 0**, so graph is **u-shaped**.
Sketch a smooth u-shaped curve through (0,6), (2,0), (3,0).

Example: Sketching y = x² - 6x + 13

**y-intercept**: (0, 13).
Complete square: **y $= (x - 3)^{2} +$ 4** → vertex at **(3, 4)** (minimum).
Since vertex is above x-axis and $a > 0$ , there are **no real roots**.
Sketch u-shaped curve with vertex (3,4) and y-intercept (0,13).

Example: Sketching y = -x² - 4x - 4

**y-intercept**: (0, -4).
Differentiate: **dy/dx $= -2x -$ 4**; set to 0 → **x = -2**.
Substitute: **y $= -(-2)^{2} - 4(-2) - 4 =$ 0** → vertex at **(-2, 0)** (maximum).
Only one root at **x = -2** (touches x-axis).
Sketch n-shaped curve with vertex (-2,0) and y-intercept (0,-4).

Example: Finding Equation from Roots

Given roots at **x = 2** and **x = 3**, and point **(0, 24)**.
Use **y $= a(x - 2)(x -$ 3)**; substitute (0,24): **24 $= a(-2)(-3) = 6a$ → $a =$ 4**.
Equation: **y $= 4(x - 2)(x -$ 3)** or **y $= 4x^{2} - 20x +$ 24**.

Example: Finding Equation from Vertex

Given vertex at **(9, -16)** and point **(2, 82)**.
Use **y $= a(x - 9)^{2} -$ 16**; substitute (2,82): **82 $= a(2-9)^{2} - 16 = 49a - 16$ → $49a = 98$ → $a =$ 2**.
Equation: **y $= 2(x - 9)^{2} -$ 16** or **y $= 2x^{2} - 36x +$ 146**.

U-shaped quadratic (a>0) with roots and y-intercept

N-shaped quadratic (a<0) with vertex on x-axis

Quadratic with no real roots (vertex above x-axis)

Finding equation from vertex and a point

Practice questions

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1.What is the shape of the graph of $y = x^{2} + 3x - 4$ ?
Easy
- AU-shaped
- BN-shaped
- CStraight line
- DS-shaped
2.The y-intercept of $y = x^{2} - 5x + 6$ is at (0, c). What is c?
Easy
- A6
- B-5
- C0
- D-6
3.How many x-intercepts can a quadratic graph have?
Easy
- A0, 1 or 2
- BAlways 2
- CAlways 1
- D0 or 2
4.The turning point of $y = (x - 3)^{2} + 5$ has coordinates:
Medium
- A(3, 5)
- B(-3, 5)
- C(3, -5)
- D(-3, -5)
5.Complete the square: $x^{2} + 6x + 11 = (x + a)^{2} + b$ . Find a and b.
Medium
- A $a = 3, b = 2$
- B $a = -3, b = 2$
- C $a = 3, b = 20$
- D $a = -3, b = 20$
6.The graph of $y = -x^{2} + 4x - 3$ has a maximum point. What is its x-coordinate?
Medium
- A2
- B-2
- C4
- D-4
7.A quadratic graph has roots at $x = 1$ and $x = 5$ . Which equation could represent it?
Medium
- A $y = (x - 1)(x - 5)$
- B $y = (x + 1)(x + 5)$
- C $y = (x - 1)(x + 5)$
- D $y = (x + 1)(x - 5)$
8.The curve $y = x^{2} - 6x + 13$ has a turning point at (3, 4). How many x-intercepts does it have?
Hard
- A0
- B1
- C2
- DCannot be determined

Summary

## Quadratic Graphs - Night Before Exam Summary ### Key Features - **General form**: $y = ax^2 + bx + c$, $a \neq 0$ - Shape: $a > 0$ → **u-shaped** (minimum turning point); $a < 0$ → **n-shaped** (maximum turning point) - The curve is a **parabola** with a vertical line of symmetry. - **y-intercept**: $(0, c)$ - **Roots** (x-intercepts): solutions to $ax^2 + bx + c = 0$ (0, 1, or 2 real roots) ### Completing the Square - Write $y = a(x - p)^2 + q$ - **Turning point** (vertex): $(p, q)$ (sign change for x!) - Example: $y = (x-3)^2 + 2$ → minimum at $(3,2)$ - $y = (x+3)^2 + 2$ → minimum at $(-3,2)$ - $a$ does not affect vertex coordinates, only shape. ### Differentiation (alternative for turning point) - $\frac{dy}{dx} = 2ax + b$ - Set $\frac{dy}{dx} = 0$ → solve for $x$ - Substitute back to find $y$. ### Sketching Steps 1. Draw axes. 2. Mark y-intercept $(0,c)$. 3. Find and mark roots (if any). 4. Determine shape (u or n). 5. Find turning point (by completing square or differentiation). 6. Draw smooth curve through intercepts and vertex. ### Finding Equation from Graph - **If vertex $(p,q)$ and another point known**: use $y = a(x-p)^2 + q$, substitute to find $a$. - **If roots $x_1, x_2$ and another point known**: use $y = a(x-x_1)(x-x_2)$, substitute to find $a$. ### Common Exam Tips - Always label intercepts and turning point coordinates. - For $y = x^2 + 10x + 14$, complete square: $(x+5)^2 - 11$, vertex $(-5,-11)$. - For $y = 6x - 0.5x^2 - 7$, rewrite as $-0.5(x-6)^2 + 11$, vertex $(6,11)$, maximum. - Axis of symmetry: $x = p$ from vertex form.

Key terms

Quadratic graph: A graph of a quadratic function $y = ax^2 + bx + c$ where $a \neq 0$. It is a smooth curve called a parabola.
Parabola: The U-shaped or n-shaped curve of a quadratic graph.
Turning point (vertex): The minimum point (if $a > 0$) or maximum point (if $a < 0$) of a quadratic graph. Coordinates can be found by completing the square: $(p, q)$ from $y = a(x-p)^2 + q$.
Roots: The x-values where the graph crosses the x-axis; solutions to $ax^2 + bx + c = 0$.
y-intercept: The point where the graph crosses the y-axis; $(0, c)$.
Completing the square: Rewriting $ax^2 + bx + c$ as $a(x-p)^2 + q$ to find the turning point.
Axis of symmetry: The vertical line through the turning point, given by $x = p$ in vertex form.

Slides

What is a Quadratic Graph?

A quadratic graph has the form $y = ax^{2} + bx + c$ , where $a \ne 0$ .
It is a smooth curve called a parabola with a vertical line of symmetry.
If $a > 0$ , the graph is u-shaped (minimum turning point).
If $a < 0$ , the graph is n-shaped (maximum turning point).

Which shape does a quadratic graph have when $a = -3$ ?

u-shaped
n-shaped
Straight line
No shape

Key Features of a Quadratic Graph

The y-intercept is at (0, c).
The x-intercepts (roots) are solutions to $ax^{2} + bx + c = 0$ .
The turning point (vertex) is the minimum or maximum point.
The graph may cross the x-axis twice, once, or not at all.

Which of these are true?

The y-intercept of $y = x^{2} - 5x + 6$ is (0, 6).
A quadratic graph always crosses the x-axis twice.
The turning point is also called the vertex.

Finding the Turning Point by Completing the Square

Write the quadratic in the form $y = a(x - p)^{2} + q$ .
The turning point is at (p, q).
Beware: the x-coordinate changes sign from the bracket.
Example: $y = (x - 3)^{2} + 2$ has minimum at (3, 2).

The turning point of $y = (x + 5)^{2} - 7$ is at (____, ____).

Finding the Turning Point by Differentiation

Differentiate $y = ax^{2} + bx + c$ to get $dy/dx = 2ax + b$ .
Set $dy/dx = 0$ and solve for x.
Substitute x back into y to find the y-coordinate.
This works for any quadratic.

For $y = x^{2} - 4x + 7$ , what is the x-coordinate of the turning point?

2
-2
4
-4

Sketching a Quadratic Graph

Draw axes and mark the y-intercept (0, c).
Find and mark the roots (solutions to $y = 0)$ .
Determine shape: u-shaped if $a > 0, n-shaped$ if $a < 0$ .
Draw a smooth curve through intercepts and mark the turning point if known.

To sketch $y = x^{2} - 5x + 6$ , the y-intercept is $(0, ____)$ and the roots are $x = 2$ and $x = ____$ . The graph $is ____$ shaped.

Finding the Equation from a Graph (Using Roots)

If roots are at x₁ and x₂, use $y = a(x -$ x₁)(x - x₂).
Substitute another known point to find a.
Example: roots at 2 and 3, point (0, 24) gives $a = 4$ .

A graph has roots at $x = 1$ and $x = 5$ , and passes through (0, 10). What is the equation?

$y = 2(x - 1)(x - 5)$
$y = (x - 1)(x - 5)$
$y = 2(x + 1)(x + 5)$
$y = (x + 1)(x + 5)$

Finding the Equation from a Graph (Using Vertex)

If vertex is at (p, q), use $y = a(x - p)^{2} + q$ .
Substitute another known point to find a.
Example: vertex (9, -16) and point (2, 82) gives $a = 2$ .

Which of these are true?

The vertex form is $y = a(x - p)^{2} + q$ .
If the vertex is (3, 4), the equation is $y = a(x - 3)^{2} + 4$ .
The value of a does not affect the shape of the graph.

Worked Example: Sketching y = x² - 5x + 6

y-intercept: (0, 6).
Factorise: $y = (x - 2)(x - 3)$ , so roots at (2,0) and (3,0).
$a = 1 > 0$ , so u-shaped.
Sketch a smooth u-shaped curve through these points.

For $y = -x^{2} - 4x - 4$ , the y-intercept is $(0, ____)$ and the turning point is at $(____, 0)$ . The graph $is ____$ shaped.

What is the turning point of $y = x^{2} + 4x - 3$ ?

(-2, -7)
(2, -7)
(-2, 7)
(2, 7)

Summary

Quadratic graphs are parabolas with a turning point and line of symmetry.
Use completing the square or differentiation to find the vertex.
Sketch by finding intercepts and the turning point.
Find the equation using roots or vertex and one other point.

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