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Year 10 Maths - 10A Solving Exponential Equations Using Logarithms

Graphs of Polynomials

20 practice questions 1 video lesson Theory + worked examples
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Theory

To sketch a polynomial, factorise it, mark the \(x\)-intercepts and \(y\)-intercept, decide whether the graph crosses or touches at each root, and use the leading term for the ends. This Year 10A maths page also covers finding an equation from a graph, inequalities and translations.

The \(x\)-intercepts of \(y = P(x)\) are the zeros of \(P(x)\). In factored form, read them straight from the factors. The \(y\)-intercept is \(P(0)\).

The power of a factor is the multiplicity of that root. At a single root the graph crosses the axis. At a double root it touches the axis and turns back. At a triple root it crosses with a flat, S-shaped bend.

The end behaviour depends only on the leading term \(ax^n\). Odd \(n\): the ends go in opposite directions. Even \(n\): both ends go the same way. A negative \(a\) flips the picture upside down.

A factor written as \((2 - x)\) is \(-(x - 2)\): it contributes a negative sign to the leading term.

This prepares you for Mathematical Methods and Specialist Mathematics in Years 11 and 12.

How a graph behaves at single, double and triple rootsThree small graphs. At a single root, from a factor (x minus a), the graph crosses the x-axis. At a double root, (x minus a) squared, it touches the axis and turns back. At a triple root, (x minus a) cubed, it crosses with a flat, S-shaped bend. (x βˆ’ a) single: crosses (x βˆ’ a)2 double: touches (x βˆ’ a)3 triple: flat crossing
The power of each factor decides what the graph does at that \(x\)-intercept.
End behaviour from the leading termFour small graphs. Odd degree with a positive leading coefficient: falls on the left, rises on the right. Odd degree, negative: rises on the left, falls on the right. Even degree, positive: rises at both ends. Even degree, negative: falls at both ends. odd degree, + odd degree, βˆ’ even degree, + even degree, βˆ’
The degree and the sign of the leading coefficient decide where the two ends of the graph go.
\[y = a(x - r)(x - s)^2(x - t)^3\]
\[\Rightarrow\ \text{cross at } r,\ \text{touch at } s,\ \text{flat cross at } t\]
y=a(x−r)(x−s)2(x−t)3
Leading termAs \(x \to -\infty\)As \(x \to \infty\)
\(ax^n\), \(n\) odd, \(a \gt 0\)\(y \to -\infty\)\(y \to \infty\)
\(ax^n\), \(n\) odd, \(a \lt 0\)\(y \to \infty\)\(y \to -\infty\)
\(ax^n\), \(n\) even, \(a \gt 0\)\(y \to \infty\)\(y \to \infty\)
\(ax^n\), \(n\) even, \(a \lt 0\)\(y \to -\infty\)\(y \to -\infty\)
New graphChange to \(y = P(x)\)
\(y = -P(x)\)reflect in the \(x\)-axis
\(y = P(-x)\)reflect in the \(y\)-axis
\(y = P(x) + c\)shift up \(c\)
\(y = P(x - h)\)shift right \(h\) (intercepts add \(h\))
\(y = kP(x)\)stretch vertically by \(k\) (same \(x\)-intercepts)
Leading term from factors. Multiply the \(x\) from each factor, with its power and sign: \((3 - x)(x + 1)^2\) has leading term \((-x)(x^2) = -x^3\).

How to sketch a polynomial

  1. Factorise fully if it is not already factored.
  2. \(x\)-intercepts: each factor equal to zero. Note the power of each factor.
  3. \(y\)-intercept: substitute \(x = 0\).
  4. Ends: find the leading term and use its degree and sign.
  5. Draw from the left end, through each intercept in order: cross at single roots, touch at double roots, flatten and cross at triple roots.

Reading graphs

  1. Matching an equation: check the intercepts, then crossing versus touching, then the ends, then the \(y\)-intercept.
  2. Finding an equation: write \(y = a(\ldots)\) from the intercepts and their multiplicities, then substitute one more point (often the \(y\)-intercept) to find \(a\).
  3. Unknown roots: if the factors contain unknowns such as \(y = a(x - m)(x - n)^2\), use the touch point for \(n\), then two points to form equations in \(a\) and \(m\).
  4. Inequalities: \(P(x) \gt 0\) where the graph is above the axis, \(P(x) \lt 0\) where it is below. Use open intervals between the zeros.
  5. Showing a touch: factorise to show the factor is squared.
Example 1 β€” Sketching a cubic in factored form
Sketch \(y = (x + 4)(x - 1)(x - 3)\), showing all intercepts.
Solution

\(x\)-intercepts: each factor equal to zero gives \(x = -4\), \(1\), \(3\). Each is a single root, so the graph crosses.

\(y\)-intercept: let \(x = 0\).

\(y\)\(=\)\((4)(-1)(-3) = \mathbf{12}\)
y=(4)(−1)(−3)=12

Ends: the leading term is \(x \cdot x \cdot x = x^3\), positive and odd, so the graph falls on the left and rises on the right.

Sketch of y = (x + 4)(x minus 1)(x minus 3)A cubic rising from bottom left, crossing the x-axis at minus 4, turning above the axis, crossing the y-axis at 12, crossing the x-axis at 1, dipping below, then crossing again at 3 and rising to the right. x y O βˆ’4 1 3 12

Draw a smooth curve through the intercepts in order. Only the intercepts are exact; the turning points are not found at this stage.

Example 2 β€” A repeated root and a negative leading term
Sketch \(y = -x^2(x + 2)(x - 3)\).
Solution

Roots: \(x^2\) gives a double root at \(x = 0\) (touch). \((x + 2)\) and \((x - 3)\) give single roots at \(-2\) and \(3\) (cross).

Ends: the leading term is \(-x^2 \cdot x \cdot x = -x^4\): even degree, negative, so both ends go down.

Check a point between the roots to see which side of the axis the graph is on.

\(x = 1:\ y\)\(=\)\(-(1)(3)(-2) = 6\)
\(x = -1:\ y\)\(=\)\(-(1)(1)(-4) = 4\)
y(1)=6,y(−1)=4

Both are above the axis, so the graph comes down to touch the origin from above and goes back up.

Sketch of y = minus x squared (x + 2)(x minus 3)A quartic that comes up from bottom left, crosses the x-axis at minus 2, rises, comes back down to touch the x-axis at the origin, rises again, then crosses at 3 and falls to the bottom right. x y O βˆ’2 3
Example 3 β€” Finding the equation from a graph
The cubic shown crosses the \(x\)-axis at \(x = -2\), touches it at \(x = 3\), and cuts the \(y\)-axis at \(9\). Find its equation.
A cubic crossing at minus 2 and touching at 3A cubic rising from bottom left, crossing the x-axis at minus 2, passing through (0, 9), turning, coming down to touch the x-axis at 3, then rising to the right. x y O (0, 9) βˆ’2 3
Solution

A crossing gives a single factor; a touch gives a squared factor. Allow for a stretch with a constant \(a\).

\(y\)\(=\)\(a(x + 2)(x - 3)^2\)

Substitute the \(y\)-intercept \((0, 9)\).

\(9\)\(=\)\(a(2)(-3)^2 = 18a\)
\(a\)\(=\)\(\mathbf{\dfrac{1}{2}}\)
y=12(x+2)(x−3)2

The equation is \(y = \dfrac{1}{2}(x + 2)(x - 3)^2\). The positive \(a\) matches the graph rising on the right.

Example 4 β€” Inequalities and translations
\(P(x) = x(x + 2)(x - 3)\).
(i) Use the graph of \(y = P(x)\) to solve \(P(x) \gt 0\).
(ii) Find the intercepts of \(y = P(x + 2)\).
Solution

(i) \(P(x) \gt 0\) where the graph is above the \(x\)-axis.

y = x(x + 2)(x minus 3) with the positive parts markedA cubic crossing the x-axis at minus 2, 0 and 3. It is above the axis between minus 2 and 0, and to the right of 3; those parts of the x-axis are marked in red. x y O βˆ’2 3
\(P(x) \gt 0\)\(\text{for}\)\(\boldsymbol{-2 \lt x \lt 0} \text{ or } \boldsymbol{x \gt 3}\)
−2<x<0 or x>3

Check \(x = -1\): \((-1)(1)(-4) = 4 \gt 0\).

(ii) Replace \(x\) with \(x + 2\): \(P(x + 2) = (x + 2)(x + 4)(x - 1)\). This is \(y = P(x)\) moved \(2\) units left.

\(x\text{-intercepts}\)\(:\)\(\mathbf{-4,\ -2,\ 1}\)
\(y\text{-intercept}\)\(:\)\(P(2) = 2(4)(-1) = \mathbf{-8}\)
P(x+2)=(x+2)(x+4)(x−1)

Common pitfalls

Crossing versus touching. A squared factor gives a touch, not a crossing. Look at the power before you draw.
Reversed factors. \((2 - x)\) changes the sign of the leading term. Count every such factor when deciding the ends.
The constant \(a\). Intercepts alone do not fix the graph: \(y = 2(x - 1)(x + 3)\) and \(y = (x - 1)(x + 3)\) share them. Use another point to find \(a\).
Translations move the opposite way. \(y = P(x + 2)\) moves the graph \(2\) units left, so each \(x\)-intercept decreases by \(2\).

Frequently asked questions

How do you sketch a polynomial graph?

Factorise it, mark the x-intercepts from the factors and the y-intercept from P(0), decide whether the graph crosses or touches at each root, and use the leading term to decide where the ends go. Then join the points smoothly.

What does a double root look like on a graph?

The graph touches the \(x\)-axis at that point and turns back, instead of crossing. It comes from a squared factor such as \((x - 2)^2\).

What is a triple root?

A root from a cubed factor. The graph crosses the x-axis there but flattens out as it crosses, making an S-shaped bend.

How do you find the end behaviour of a polynomial?

Look at the leading term. If the degree is odd, the ends go in opposite directions; if even, the same direction. A positive leading coefficient means the right-hand end goes up.

How do you find the equation of a polynomial from its graph?

Write a factor for each x-intercept, squared if the graph touches and cubed if it flattens while crossing. Put a constant a in front, then substitute another point on the graph, usually the y-intercept, to find a.