Quadratics
Quadratics are polynomials of degree 2.
They come in many forms, but always contain a squared term.
Quadratic expression:
Quadratic equation:
Quadratic function:
Quadratic graph:
Quadratic inequation:
Quadratic mapping:
Quadratic Graphs
Quadratic graphs have a distinctive U‑shape called a parabola.
Positive parabolas smile:
\(y = ax^2\)
Negative parabolas frown:
\(y = -ax^2\)
Drawing Parabolas of the Form \(y = ax^2\)
Pick values for \(x\) and put them into a table.
Work out the corresponding values of \(y\).
Plot these points and join them with a smooth curve.
Complete the table of values and sketch the graph of the equation \(\; y = x^2\)
Substituting into a Function
Problem:
When \( x = -2 \), find the value of \( y \) for the function
\[
y = x^2
\]
Step-by-step
Substitute \( x = -2 \) into the formula: \[ y = x^2 \]
\[ y = (-2)^2 \]
\[ y = 4 \]
Conclusion
So the point \( (-2, 4) \) lies on the curve.
Table of Values
| x | y = x^2 |
|---|---|
| -2 | \( (-2)^2 = 4 \) |
| -1 | \( (-1)^2 = 1 \) |
| 0 | \( 0^2 = 0 \) |
| 1 | \( 1^2 = 1 \) |
| 2 | \( 2^2 = 4 \) |
Plotting these points and joining with a smooth curve gives:
Graph of \(y = x^2\)
Notice how the graph is symmetrical!
Complete the table of values and sketch the parabola of the equation \(\; y = -5x^2\)
Table of Values
| x | y = -5x² |
|---|---|
| -2 | \( -5(-2)^2 = -20 \) |
| -1 | \( -5(-1)^2 = -5 \) |
| 0 | \( -5(0)^2 = 0 \) |
| 1 | \( -5(1)^2 = -5 \) |
| 2 | \( -5(2)^2 = -20 \) |
Plotting these points and joining with a smooth curve gives:
Graph of \(y = -5x^2\)
Again, notice how the graph is symmetrical!
Working Backwards
Find the equation of the following parabola of the form \(y = ax^2\)
Parabola with point (2,1)
The graph is of the form \(y = ax^2\). The given coordinate is \((2,1)\). So \(x = 2\) and \(y = 1\) are on the curve.
Substitute and solve:
Parabolas of the form \(y = a(x-b)^2\)
Complete the table of values and sketch the equation \(y = (x - 2)^2\)
Table of Values for \( y = (x - 2)^2 \)
| x | y |
|---|---|
| -1 | \[ y = (-1 - 2)^2 = (-3)^2 = 9 \] |
| 0 | \[ y = (0 - 2)^2 = (-2)^2 = 4 \] |
| 1 | \[ y = (1 - 2)^2 = (-1)^2 = 1 \] |
| 2 | \[ y = (2 - 2)^2 = 0^2 = 0 \] |
| 3 | \[ y = (3 - 2)^2 = 1^2 = 1 \] |
Plotting these points and joining with a smooth curve gives:
Graph of \(y = (x - 2)^2\)
The graph is symmetrical when \(x = 2\).
The turning point is \((2,0)\).
The axis of symmetry is the value of \(b\) in the equation \(y = a(x - b)^2\).
Find the equation of the following parabola of the form \(y = a(x - b)^2\)
Parabola with point (2,3)
The graph is of the form \(y = a(x - b)^2\). The given coordinate is \((2,3)\). So \(x = 2\) and \(y = 3\) are on the curve.
Substitute and solve:
\[ \; y = a(x - b)^2 \] \[ \;\;\;\; y = a(x - 1)^2 \] \[ \;\;\;\; 3 = a(2 - 1)^2 \] \[ \;\;\;\; 3 = a \times (-1)^2 \] \[ \;\;\;\; a = 3 \] \[ \;\;\;\; y = 3(x - 1)^2 \]Parabolas of the form \(y = a(x-b)^2 + c\)
Complete the table of values and sketch the graph of the equation \(y = -2(x + 3)^2 + 2\)
Table of Values for \( y = -2(x + 3)^2 + 2 \)
| x | y |
|---|---|
| -5 | \[ y = -2(-5 + 3)^2 + 2 = -2(-2)^2 + 2 = -8 + 2 = -6 \] |
| -4 | \[ y = -2(-4 + 3)^2 + 2 = -2(-1)^2 + 2 = -2 + 2 = 0 \] |
| -3 | \[ y = -2(-3 + 3)^2 + 2 = -2(0)^2 + 2 = 2 \] |
| -2 | \[ y = -2(-2 + 3)^2 + 2 = -2(1)^2 + 2 = -2 + 2 = 0 \] |
| -1 | \[ y = -2(-1 + 3)^2 + 2 = -2(2)^2 + 2 = -8 + 2 = -6 \] |
| 0 | \[ y = -2(0 + 3)^2 + 2 = -2(3)^2 + 2 = -18 + 2 = -16 \] |
| 1 | \[ y = -2(1 + 3)^2 + 2 = -2(4)^2 + 2 = -32 + 2 = -30 \] |
Graph of \(y = -2(x+3)^2 + 2\)
Notice that the axis of symmetry is \(x = -3\).
Working Backwards
Find the equation of the following parabola of the form \(y = a(x - b)^2 + c\)
The graph is of the form \(y = a(x - b)^2 + c\). The given coordinate is \((-3,-2)\). So \(x = -3\) and \(y = -2\) are on the curve.
From the graph, \(b = -2\) since it is the axis of symmetry.
Substitute \(x=-3\), \(y=-2\), \(b=-2\):
\[ \;\;\;\; y = a(x - b)^2 + c \] \[ \;\;\;\; -2 = a(-3 - (-2))^2 + c \] \[ \;\;\;\; -2 = a((-1))^2 + c \] \[ \;\;\;\; -2 = a + c \]The point \((-2,-5)\) is also on the curve:
\[ \;\;\;\; y = a(x - b)^2 + c \] \[ \;\;\;\; -5 = a(-2 - (-2))^2 + c \] \[ \;\;\;\; -5 = a(0)^2 + c \] \[ \;\;\;\; -5 = c \]So \(c = -5\).
Substitute into \(-2 = a + c\):
\[ -2 = a + c \] \[ \;\;\;\;\; -2 = a - 5 \] \[ \;\;\;\;\; 5 - 2 = a \] \[ \;\;\;\;\;\;\;\; 3 = a \]Substituting \(a\), \(b\), and \(c\) into the equation:
\[ y = 3(x + 2)^2 - 5 \]Axis of Symmetry
This is a quadratic in completed square form.
Find the axis of symmetry of \(y = x^2 + 3x + 2\)
\[ \text{Completing the square} \] \[ y = \left(x + \frac{3}{2}\right)^2 \; + 2 - \frac{9}{4} \] \[ y = \left(x + \frac{3}{2}\right)^2 \; + \frac{8}{4} - \frac{9}{4} \] \[ y = \left(x + \frac{3}{2}\right)^2 \; - \frac{1}{4} \]The axis of symmetry is \(x = -\tfrac{3}{2}\).
Parabolas of the form \(y = ax^2 + bx + c\)
Sketch the parabola \(y = 2x^2 + 3x - 2\)
Table of Values for \( y = 2x^2 + 3x - 2 \)
| x | y |
|---|---|
| -2 | \[ y = 2(-2)^2 + 3(-2) - 2 = 8 - 6 - 2 = 0 \] |
| -1 | \[ y = 2(-1)^2 + 3(-1) - 2 = 2 - 3 - 2 = -3 \] |
| 0 | \[ y = 2(0)^2 + 3(0) - 2 = -2 \] |
| 1 | \[ y = 2(1)^2 + 3(1) - 2 = 2 + 3 - 2 = 3 \] |
| 2 | \[ y = 2(2)^2 + 3(2) - 2 = 8 + 6 - 2 = 12 \] |
Graph of \(y = 2x^2 + 3x - 2\)
Turning Points
Positive parabolas have a minimum turning point.
Find the turning point of the quadratic
Graph of \(y = x^2 + 3x + 2\)
The turning point occurs on the axis of symmetry.
\[ y = x^2 + 3x + 2 \] \[ \text{Completing the square} \] \[ y = \left(x + \frac{3}{2}\right)^2 \; + 2 - \frac{9}{4} \] \[ y = \left(x + \frac{3}{2}\right)^2 \; + \frac{8}{4} - \frac{9}{4} \] \[ y = \left(x + \frac{3}{2}\right)^2 \; - \frac{1}{4} \] \[ \text{Giving turning point co-ordinates } \] \[ \left( -\frac{3}{2},\; -\frac{1}{4} \right) \]Negative parabolas have a maximum turning point.
Roots
A root of an equation is a value that will satisfy the equation when its expression is set to zero.
For example: \[ 0 = x^2 + 2x - 3 \]
The maximum number of roots possible is the same as the degree of the polynomial, so a quadratic can have a maximum of two roots. Not all quadratics have roots.
To find the roots of a quadratic, sketch the graph and see where it cuts the \(x\)-axis.
Or:
Set \(y = 0\) and factorise (if possible).
Graph of \(y = x^2 + 2x - 3\)
From the graph, the equation \(y = x^2 + 2x - 3\) has roots \(x = -3\) and \(x = 1\).
This is the same as setting \(y = 0\) and factorising:
\[ y = x^2 \; + \; 2x \; - 3 \] \[ 0 = x^2 \; + \; 2x \; - 3 \] \[ 0 = (x - 1)(x + 3) \]Either bracket can equal 0, so both must be considered:
\[ \;\;\;\;\;\;\text{so either} \] \[ (x - 1) = 0 \;\;\;\;\;\text{or}\;\;\;\;\; (x + 3) = 0 \] \[ \;\; x - 1 = 0 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; x + 3 = 0 \] \[ \text{giving} \] \[ \;\;\;\;\;\;\;\; x = 1 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; x = -3 \] \[ \text{The roots are } x = 1 \text{ and } x = -3 \] \[ \text{with co-ords } (1,0),\, (-3,0) \]Sketching Parabolas
To sketch the graph
\(y = ax^2 + bx + c\)
Remember: shape, roots, turning point, \(y\)-intercept.
Sketch \(y = x^2 - 2x - 3\).
This will be a U shape, since \(a = 1\).
It will cut the \(y\)-axis at \((0,-3)\).
Sketch of \(y = x^2 - 2x - 3\)
Sketch \(y = 3 - 2x - x^2\).
This will be a ∩ shape, since \(a = -1\).
It will cut the \(y\)-axis at \((0,3)\).
\[ \text{Turning point occurs on axis of symmetry} \] \[ \text{When } x = -1, \] \[ \;\;\;\;\;\;\;\; y = 3 - 2x - x^2 \] \[ \;\;\;\;\;\;\;\; y = 3 - 2(-1) - (-1)^2 \] \[ \;\;\;\;\;\;\;\; y = 3 - (-2) - 1 \] \[ \;\;\;\;\;\;\;\; y = 4 \] \[ \text{Turning point } (-1, 4) \]
Sketch of \(y = 3 - 2x - x^2\)
Quadratic Equations
Standard form quadratic equations are of the form:
To find the solution of a quadratic equation:
- Rewrite the expression in standard quadratic form.
- Factorise if you can.
{Remember to look for common factors and the difference of two squares.}
Or use the quadratic formula:
Solve \(3x - 6x^2 = 0\).
\[ 3x - 6x^2 = 0 \] \[ 3x(1 - 2x) = 0 \] \[ so \] \[ 3x = 0 \;\;\;\;\; or \;\;\;\;\; 1 - 2x = 0 \] \[ x = 0 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; 2x = 1 \] \[ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; x = \frac{1}{2} \] \[ x = 0,\;\;\; x = \frac{1}{2} \]Solve \(49 - 9x^2 = 0\).
\[ 49 - 9x^2 = 0 \] \[ 7^2 - (3x)^2 = 0 \] \[ (7 + 3x)(7 - 3x) = 0 \] \[ so \] \[ 7 + 3x = 0 \;\;\;\;\;\text{or}\;\;\;\;\; 7 - 3x = 0 \] \[ 3x = -7 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; 3x = 7 \] \[ x = -\frac{7}{3} \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; x = \frac{7}{3} \] \[ \text{solutions} \] \[ x = -\frac{7}{3},\;\;\; x = \frac{7}{3} \]Solve \(15x^2 - x - 6 = 0\).
\[ 15x^2 - x - 6 = 0 \] \[ (3x - 2)(5x + 3) = 0 \] \[ (3x - 2) = 0 \;\;\;\;\;\text{or}\;\;\;\;\; (5x + 3) = 0 \] \[ 3x - 2 = 0 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; 5x + 3 = 0 \] \[ 3x = 2 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; 5x = -3 \] \[ x = \frac{2}{3} \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; x = -\frac{3}{5} \] \[ \text{solutions} \] \[ x = \frac{2}{3},\;\;\; x = -\frac{3}{5} \]Solve \(15x^2 - x + 1 = 7\).
\[ 15x^2 - x + 1 = 7 \] \[ 15x^2 - x - 6 = 0 \] \[ (3x - 2)(5x + 3) = 0 \] \[ (3x - 2) = 0 \;\;\;\;\;\text{or}\;\;\;\;\; (5x + 3) = 0 \] \[ 3x - 2 = 0 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; 5x + 3 = 0 \] \[ 3x = 2 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; 5x = 3 \] \[ x = \frac{2}{3} \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; x = \frac{3}{5} \]Quadratic Formula
If the quadratic does not factorise, try the quadratic formula:
Solve \(2 + 4x - 5x^2 = 0\).
Give your answer as a surd.
\[ \text{Re-write in standard form } -5x^2 + 4x + 2 = 0 \] \[ \;\;\;\;\text{so } a = -5,\; b = 4,\; \text{and } c = 2 \] \[ \text{Use } x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] \[ x = \frac{-4 \pm \sqrt{4^2 - 4(-5)(2)}}{2(-5)} \] \[ x = \frac{-4 \pm \sqrt{16 - (-40)}}{-10} \] \[ x = \frac{-4 + \sqrt{56}}{-10} \;\;\;\text{and}\;\;\; x = \frac{-4 - \sqrt{56}}{-10} \] \[ x = \frac{-4 + 2\sqrt{14}}{-10} \;\;\;\text{and}\;\;\; x = \frac{-4 - 2\sqrt{14}}{-10} \] \[ x = \frac{-2(2 - \sqrt{14})}{-10} \;\;\;\text{and}\;\;\; x = \frac{-2(2 + \sqrt{14})}{-10} \] \[ x = \frac{2 - \sqrt{14}}{5} \;\;\;\text{and}\;\;\; x = \frac{2 + \sqrt{14}}{5} \]Find the roots of \(2 + 4x - 5x^2\).
Give your answer correct to two decimal places.
\[ \text{Re-write in standard form } -5x^2 + 4x + 2 = 0 \] \[ \;\;\;\;\text{so } a = -5,\; b = 4,\; \text{and } c = 2 \]\[ \text{Use } x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] \[ x = \frac{-4 \pm \sqrt{4^2 - 4(-5)(2)}}{2(-5)} \] \[ x = \frac{-4 \pm \sqrt{16 - (-40)}}{-10} \] \[ x = \frac{-4 + \sqrt{56}}{-10} \;\;\;\text{and}\;\;\; x = \frac{-4 - \sqrt{56}}{-10} \] \[ x = \frac{-4 + 2\sqrt{14}}{-10} \;\;\;\text{and}\;\;\; x = \frac{-4 - 2\sqrt{14}}{-10} \] \[ x = \frac{-2(2 - \sqrt{14})}{-10} \;\;\;\text{and}\;\;\; x = \frac{-2(2 + \sqrt{14})}{-10} \] \[ x = \frac{2 - \sqrt{14}}{5} \;\;\;\text{and}\;\;\; x = \frac{2 + \sqrt{14}}{5} \] \[ x = -0.348\;\;\;\text{and}\;\;\; x = 1.148 \] \[ \text{Roots } x = -0.35 \;\;\text{and}\;\; x = 1.15 \] \[ \text{Roots have co-ordinates } (-0.35, 0) \text{ and } (1.15, 0) \]
Discriminant
The discriminant of a quadratic \(ax^2 + bx + c = 0\) is:
It tells us how many real roots the quadratic has:
State the nature of the roots of the equation \( y = x^2 + 3x + 4 \)
Since \(b^2 - 4ac \lt 0\), there are No Real Roots.
No real roots
State the nature of the roots of the equation \( y = x^2 + 6x + 9 \)
Since \(b^2 - 4ac = 0\), there is one real repeated root.
The root is at the coordinate (-3, 0).
Two distinct real roots
State the nature of the roots of the equation \( y =12 +2x - 2x^2 \)
Since \(b^2 - 4ac > 0\), there are two real , distinct roots.
The roots are at the coordinates (-2, 0) and (3, 0).
Two distinct, real roots
Working Backwards
The roots of \((x - 1)(x + k) = -4\) are equal. Find the value of \(k\).
First multiply out the brackets:
\[ \;(x - 1)(x + k) = -4 \] \[ x^2 - x + kx - k = -4 \] \[ x^2 + x(k - 1) - k = -4 \] \[ x^2 + x(k - 1) + 4 - k = 0 \] \[ x^2 + (k - 1)x + (4 - k) = 0 \] \[ \text{Discriminant} \] \[ b^2 - 4ac = (k - 1)^2 - 4 \cdot 1 \cdot (4 - k) \] \[ \;\;\;\;\;\; = (k - 1)^2 - 4(4 - k) \] \[ \text{Roots are equal, so discriminant } = 0 \] \[ (k - 1)^2 - 4(4 - k) = 0 \] \[ \Rightarrow k^2 - 2k + 1 - 16 + 4k = 0 \] \[ \Rightarrow k^2 + 2k - 15 = 0 \] \[ \Rightarrow (k - 5)(k + 3) = 0 \] \[ \Rightarrow k = 5 \;\;\text{or}\;\; k = -3 \]Tangency
A tangent to a curve touches it at exactly one point. To test for tangency, set the two functions equal and examine the discriminant.
\[ \text{Find the point of tangency:} \] \[ x = \frac{-b}{2a} = \frac{-4}{2} = -2 \] \[ y = 2(-2) + 5 = 1 \] \[ \text{The line touches the curve at } (-2,\,1). \]
