Quadratic Graphs

Quadratic Graphs

A LevelAQAEdexcelOCRAQA 2022OCR 2022

Quadratic Graphs

You will need to be able to identify and sketch quadratic graphs.

Make sure you are happy with the following topics before continuing.

A Level AQA Edexcel OCR

Quadratic Graph Shapes

Quadratic graphs have the general form

\textcolor{red}{a} x^2 + \textcolor{blue}{b}x + \textcolor{limegreen}{c}

These can be either u-shaped (if \textcolor{red}{a} is positive) or n-shaped (if \textcolor{red}{a} is negative)

Some examples of quadratic graphs can bee seen below:

A LevelAQAEdexcelOCR

Sketching Quadratic Graphs

To sketch quadratic graphs, you need to follow these steps:

Step 1: Decide whether the graph is an n-shape or a u-shape.

  • u-shaped if the coefficient of x^2 is positive
  • n-shaped if the coefficient of x^2 is negative

Step 2: Find the intersections with the x-axis (set y=0) and y-axis (set x=0)

Step 3: Find the maximum (if it is an n-shaped graph) or minimum (if it is a u-shaped graph) using symmetry or completing the square.

Step 4: Sketch the graph and label all the important points.

 

Example: Sketch the graph of y = 2x^2 + 3x - 2

Step 1: The coefficient of x^2 is positive, so the graph is u-shaped.

Step 2: When y=0,

\begin{aligned} 2x^2 + 3x - 2 &= 0 \\ (2x-1)(x+2) &= 0 \end{aligned}

x = \dfrac{1}{2}  or  x = -2

Hence, the curve crosses the x-axis at \textcolor{red}{\left( \dfrac{1}{2}, 0 \right)} and \textcolor{red}{(-2,0)}

When x = 0,

y = 2(0)^2 + 3(0) - 2 = -2

Hence, the curve crosses the y-axis at \textcolor{red}{(0,-2)}

Note: You can often read off the value the graph crosses the y-axis at (called the y-intercept) from the equation of a quadratic. The y-intercept of y=ax^{2}+bx+c is always c.

Step 3: The minimum value will be halfway between \left(\dfrac{1}{2}, 0 \right) and (-2,0):

\left( \dfrac{1}{2} - 2 \right) \div 2 = - \dfrac{3}{4}

So, the minimum value is at x = - \dfrac{3}{4}

Therefore, the minimum is

y = 2 \left(- \dfrac{3}{4} \right)^2 + 3 \left(- \dfrac{3}{4} \right) - 2 = - \dfrac{25}{8}

So, the graph has a minimum at the point \textcolor{red}{\left(- \dfrac{3}{4}, - \dfrac{25}{8} \right)}

Step 4: Sketch the graph (sketching doesn’t need to be to scale) and label all the important points.

A LevelAQAEdexcelOCR

Completing the Square to Sketch Graphs

You can complete the square to help sketch quadratic curves.

Example: Sketch the curve of y = 3x^2 - 12x + 7

Firstly, complete the square:

y = 3x^2 - 12x + 7 = 3(x-2)^2 - 5

When y=0,

\begin{aligned} 3(x-2)^2 - 5 &=0 \\ (x-2)^2 &= \dfrac{5}{3} \\ x - 2 &= \pm \sqrt{\dfrac{5}{3}} \\ x &= 2 \pm \dfrac{\sqrt{15}}{3} \end{aligned}

So, the curve crosses the x-axis at \textcolor{red}{\left(2-\dfrac{\sqrt{15}}{3},0 \right)} and \textcolor{red}{\left(2+\dfrac{\sqrt{15}}{3},0 \right)}

When x=0,

y = 3(0)^2 - 12(0) + 7 = 7

So, the curve crosses the y-axis at \textcolor{red}{(0,7)}

(x-2)^2 \geq 0 (since a square number is never less than 0), therefore the minimum value occurs at x-2=0 \Rightarrow x = 2, and therefore the minimum is

y = 3(2-2)^2 - 5

So, the minimum is at the point \textcolor{red}{(2, -5)}

Then, you have enough information to sketch the graph.

Note: You can read off the vertex of a graph from completed square form. The vertex of y=a(x+b)^{2}+c is always at (-b,c).

A LevelAQAEdexcelOCR

Functions with no Real Roots

Some functions have no real roots – i.e. its graph doesn’t cross the x-axis. You can complete the square and use this to see if quadratics have real roots, or not.

Example: f(x) = x^2 + 2x + 6. Determine whether f(x) has any real roots.

Complete the square:

f(x) = x^2 + 2x + 6 = \textcolor{red}{(x+1)^2} + \textcolor{blue}{5}

\textcolor{red}{(x+1)^2} is never less than 0, therefore the minimum of the curve of f(x) is at \textcolor{blue}{5}.

Hence, f(x) cannot be negative and will not cross the x-axis at any point.

Therefore, f(x) has no real roots.

 

Note: If the coefficient of x^2 is negative, you can do a similar thing to see if f(x) has real roots, by checking if f(x) is ever positive.

A LevelAQAEdexcelOCR

Example Questions

The coefficient of x^2 is negative, so the graph is n-shaped.

When x = 0, y = 6

When y = 0,

\begin{aligned} 6 - x - x^2 &= 0 \\ (2-x)(x+3) &= 0 \end{aligned}

x = 2 or x = -3

The maximum value is halfway between -3 and 2. So the maximum is at x = - \dfrac{1}{2} and y = 6 - \left(- \dfrac{1}{2} \right) - \left(- \dfrac{1}{2} \right)^2 = \dfrac{25}{4}

So, we have enough information to plot the graph:

The coefficient of x^2 is positive, so there will be a minimum.

 

\begin{aligned} f(x) &= 2x^2 + 16x - 10 \\ &= 2 \left(x+\dfrac{16}{2 \times 2} \right)^2 + \left(-10 - 2 \left(\dfrac{16}{2 \times 2}\right)^2 \right) \\ &= 2(x+4)^2 - 42 \end{aligned}

 

Hence, the minimum occurs at (x+4) = 0 \Rightarrow x = -4

 

So, the minimum is

f(-4) = 2(-4+4)^2 - 42 = -42

 

Hence, the graph has a minimum at the point (-4, -42)

\begin{aligned} f(x) &= -2x^2 - 8x - 12 \\ &= -(2x^2 + 8x + 12) \end{aligned}

 

Complete the square:

2x^2 + 8x + 12 \\ = 2 \left(x+\dfrac{8}{2 \times 2} \right)^2 + \left(12 - 2\left(\dfrac{8}{2 \times 2}\right)^2 \right) \\ = 2(x+2)^2 + 4

 

So, f(x) = - 2(x+2)^2 - 4

 

(x+2)^2 is never less than 0, therefore the maximum of the curve of f(x) is at -4.

Hence, f(x) cannot be positive and will not cross the x-axis at any point.

Therefore, the function has no real roots.

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