Year 12 Maths - Methods (Unit 3 & Unit 4) Polynomial functions

Quadratic functions

Q: What are the three forms of a quadratic function?

A quadratic can be written in general form y equals a x squared plus b x plus c, in turning-point (vertex) form y equals a times (x minus h) squared plus k where (h, k) is the turning point, and in factored (intercept) form y equals a times (x minus p)(x minus q) where p and q are the x-intercepts. The same parabola is described by all three; you choose the form that shows the feature you need.

Q: How do you find the turning point of a parabola?

The axis of symmetry is x equals minus b over 2a; substitute this x value back into the rule to get the y-coordinate of the turning point. In turning-point form y equals a times (x minus h) squared plus k the turning point is read straight off as (h, k).

Q: What is completing the square used for?

Completing the square rewrites y equals a x squared plus b x plus c in the turning-point form y equals a times (x minus h) squared plus k. This shows the turning point directly and describes how the parabola is transformed from the basic curve y equals x squared.

Q: What does the discriminant tell you?

The discriminant is delta equals b squared minus 4 a c. If delta is greater than zero there are two real solutions and the parabola cuts the x-axis twice; if delta equals zero there is one solution and the parabola touches the x-axis; if delta is less than zero there are no real solutions and the parabola does not cross the x-axis.

Q: How do you tell if a parabola is concave up or concave down?

The sign of the coefficient a decides it. If a is greater than zero the parabola is concave up and has a minimum turning point; if a is less than zero the parabola is concave down and has a maximum turning point. A larger value of the absolute value of a makes the parabola narrower.

Q: When do you use the quadratic formula?

Use the quadratic formula x equals minus b plus or minus the square root of (b squared minus 4 a c), all over 2a, to solve a x squared plus b x plus c equals zero when the expression does not factorise neatly. The part under the square root is the discriminant, so it also tells you how many real solutions there are.

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Theory

A quadratic function \(y=ax^{2}+bx+c\) (with \(a\neq 0\)) graphs as a parabola. It can also be written in turning-point form \(y=a(x-h)^{2}+k\) or factored form \(y=a(x-p)(x-q)\). The sign of \(a\) fixes the concavity, the axis of symmetry is \(x=-\dfrac{b}{2a}\), and the discriminant \(\Delta=b^{2}-4ac\) counts the \(x\)-intercepts.

A quadratic function is any function of the form \(y=ax^{2}+bx+c\) with \(a\neq 0\); its graph is a parabola. The coefficient \(a\) controls the shape: when \(a>0\) the parabola is concave up and has a minimum turning point, and when \(a<0\) it is concave down with a maximum. The constant \(c\) is the \(y\)-intercept, since setting \(x=0\) gives \(y=c\).

The same parabola can be written in three equivalent forms. The general form \(y=ax^{2}+bx+c\) is handy for the \(y\)-intercept and the discriminant; the turning-point (vertex) form \(y=a(x-h)^{2}+k\) shows the turning point \((h,k)\) directly; and the factored (intercept) form \(y=a(x-p)(x-q)\) shows the \(x\)-intercepts \(x=p\) and \(x=q\). Completing the square converts the general form to turning-point form.

Every parabola is symmetric about the vertical line through its turning point, the axis of symmetry \(x=-\dfrac{b}{2a}\). The number of \(x\)-intercepts is decided by the discriminant \(\Delta=b^{2}-4ac\): two if \(\Delta>0\), one (a touch) if \(\Delta=0\), and none if \(\Delta<0\). When factorising is awkward, the quadratic formula \(x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) solves \(ax^{2}+bx+c=0\).

Key idea. One parabola, three forms. Read the \(y\)-intercept from \(y=ax^{2}+bx+c\), the turning point from \(y=a(x-h)^{2}+k\), and the \(x\)-intercepts from \(y=a(x-p)(x-q)\). The sign of \(a\) gives concavity; \(\Delta=b^{2}-4ac\) gives the number of \(x\)-intercepts.

\(y=x^{2}-2x-3=(x-1)^{2}-4=(x-3)(x+1)\): intercepts, turning point \((1,-4)\) and axis \(x=1\)

The discriminant \(\Delta=b^{2}-4ac\) counts the \(x\)-intercepts: two \((\Delta>0)\), one \((\Delta=0)\) or none \((\Delta<0)\)

The three equivalent forms of a quadratic:

\[y=ax^{2}+bx+c \qquad y=a(x-h)^{2}+k \qquad y=a(x-p)(x-q)\]

y = a x^{2} + b x + c

The axis of symmetry, and the \(x\)-coordinate of the turning point:

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

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

The discriminant, which counts the real solutions of \(ax^{2}+bx+c=0\):

\[\Delta=b^{2}-4ac\]

Δ = b^{2} - 4 a c

The quadratic formula, for solving \(ax^{2}+bx+c=0\):

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

x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}

Reading the graph. The \(y\)-intercept is \(c\); the turning point is \(\left(-\dfrac{b}{2a},\,f\!\left(-\dfrac{b}{2a}\right)\right)\); and \(\Delta=b^{2}-4ac\) gives \(2\), \(1\) or \(0\) \(x\)-intercepts as \(\Delta\) is positive, zero or negative.

How to sketch and analyse a parabola

Read off \(a\), \(b\), \(c\) and the concavity. If \(a>0\) the parabola is concave up (minimum); if \(a<0\) it is concave down (maximum).
Find the \(y\)-intercept. Put \(x=0\): the \(y\)-intercept is \(c\).
Find the \(x\)-intercepts. Solve \(ax^{2}+bx+c=0\) by factorising, or by the quadratic formula; use \(\Delta=b^{2}-4ac\) to check how many there are.
Find the turning point. The axis of symmetry is \(x=-\dfrac{b}{2a}\); substitute this \(x\) back in to get the \(y\)-value, or read \((h,k)\) from \(y=a(x-h)^{2}+k\).
Plot and draw. Mark the intercepts and turning point, then join them with a smooth, symmetric curve.

Vertex-form shortcut. If a rule is given as \(y=a(x-h)^{2}+k\), the turning point is \((h,k)\) with no calculation — watch the sign, since \(y=(x-3)^{2}+2\) has turning point \((3,2)\).

Example 1 — sketch from turning-point form

Sketch \(y=(x-2)^{2}-1\), showing the turning point and all intercepts.

Solution

Turning point and concavity — read from \(y=a(x-h)^{2}+k\):

\(a=1>0\)	\(\Rightarrow\)	concave up (minimum)
turning point	\(=\)	\((2,-1)\)

\(y\)-intercept — substitute \(x=0\):

\(y\)

\(=\)

\((0-2)^{2}-1=3\)

\(x\)-intercepts — set \(y=0\):

\((x-2)^{2}-1\)	\(=\)	\(0\)
\((x-2)^{2}\)	\(=\)	\(1\)
\(x-2\)	\(=\)	\(\pm 1\)
\(x\)	\(=\)	\(1 \text{ or } 3\)

\(\therefore\) minimum \((2,-1)\); \(x\)-intercepts \((1,0),(3,0)\); \(y\)-intercept \((0,3)\)

y = {(x - 2)}^{2} - 1

Example 2 — completing the square

Write \(y=x^{2}+6x+5\) in turning-point form and state the turning point.

Solution

Halve the coefficient of \(x\) and square it — here \(\left(\dfrac{6}{2}\right)^{2}=9\):

\(y\)	\(=\)	\(x^{2}+6x+5\)
	\(=\)	\((x^{2}+6x+9)-9+5\)

Write the perfect square and simplify:

\(y\)

\(=\)

\((x+3)^{2}-4\)

Read the turning point from \(y=a(x-h)^{2}+k\) — here \(h=-3\), \(k=-4\):

turning point

\(=\)

\((-3,-4)\)

\(\therefore\) \(y=(x+3)^{2}-4\); minimum turning point \((-3,-4)\)

y = {(x + 3)}^{2} - 4

Example 3 — using the discriminant

(a) How many \(x\)-intercepts does \(y=2x^{2}+3x+5\) have? (b) For what values of \(k\) does \(y=x^{2}+kx+9\) touch the \(x\)-axis?

Solution

(a)

Evaluate \(\Delta=b^{2}-4ac\) with \(a=2,\ b=3,\ c=5\):

\(\Delta\)	\(=\)	\(3^{2}-4(2)(5)\)
	\(=\)	\(9-40=-31\)

\(\Delta<0\), so there are no \(x\)-intercepts

(b)

Touching the \(x\)-axis means one solution, so set \(\Delta=0\) with \(a=1,\ b=k,\ c=9\):

\(k^{2}-4(1)(9)\)	\(=\)	\(0\)
\(k^{2}\)	\(=\)	\(36\)
\(k\)	\(=\)	\(\pm 6\)

\(\therefore\) \(k=6\) or \(k=-6\)

Δ = k^{2} - 36 = 0

Example 4 — turning point and quadratic formula

For \(y=2x^{2}-8x+5\), find the turning point and interpret it, then find the \(x\)-intercepts.

Solution

Axis of symmetry — use \(x=-\dfrac{b}{2a}\) with \(a=2,\ b=-8\):

\(x\)

\(=\)

\(-\dfrac{-8}{2(2)}=2\)

Substitute \(x=2\) to get the \(y\)-value of the turning point:

\(y\)	\(=\)	\(2(2)^{2}-8(2)+5\)
	\(=\)	\(8-16+5=-3\)

Since \(a=2>0\) the parabola is concave up, so the turning point is a minimum:

turning point

\(=\)

\((2,-3)\), a minimum

\(x\)-intercepts — the quadratic formula with \(\Delta=(-8)^{2}-4(2)(5)=24\):

\(x\)

\(=\)

\(\dfrac{8\pm\sqrt{24}}{4}=2\pm\dfrac{\sqrt{6}}{2}\)

\(\therefore\) minimum \((2,-3)\); \(x\)-intercepts \(x=2\pm\dfrac{\sqrt{6}}{2}\)

x = 2 \pm \frac{\sqrt{6}}{2}

Common pitfalls

The turning point sign is opposite the bracket. In \(y=a(x-h)^{2}+k\) the turning point is \((h,k)\), so \(y=(x+3)^{2}-4\) has turning point \((-3,-4)\), not \((3,-4)\).

Do not forget the minus in the axis of symmetry. The axis is \(x=-\dfrac{b}{2a}\); for \(y=2x^{2}-8x+5\) this is \(-\dfrac{-8}{4}=2\), a positive value.

\(\Delta<0\) means no real \(x\)-intercepts, not one. Only \(\Delta=0\) gives a single (repeated) solution where the parabola touches the axis; \(\Delta<0\) means the parabola never crosses it.

Frequently asked questions

What are the three forms of a quadratic function?

General form \(y=ax^{2}+bx+c\), turning-point (vertex) form \(y=a(x-h)^{2}+k\) with turning point \((h,k)\), and factored (intercept) form \(y=a(x-p)(x-q)\) with \(x\)-intercepts \(p\) and \(q\). Choose the form that shows the feature you need.

How do you find the turning point of a parabola?

The axis of symmetry is \(x=-\dfrac{b}{2a}\); substitute this \(x\) back into the rule for the \(y\)-coordinate. In turning-point form \(y=a(x-h)^{2}+k\) the turning point is simply \((h,k)\).

What is completing the square used for?

It rewrites \(y=ax^{2}+bx+c\) as \(y=a(x-h)^{2}+k\), which shows the turning point directly and how the parabola is transformed from \(y=x^{2}\).

What does the discriminant tell you?

\(\Delta=b^{2}-4ac\). If \(\Delta>0\) there are two \(x\)-intercepts; if \(\Delta=0\) the parabola touches the axis once; if \(\Delta<0\) there are no real solutions.

How do you tell if a parabola is concave up or concave down?

The sign of \(a\) decides it: \(a>0\) gives concave up with a minimum, and \(a<0\) gives concave down with a maximum. A larger \(|a|\) makes it narrower.

When do you use the quadratic formula?

Use \(x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) to solve \(ax^{2}+bx+c=0\) when it does not factorise neatly; the term under the root is the discriminant.

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Determining the rule for a parabola

Quadratic functions

📖 Theory

How to sketch and analyse a parabola

Common pitfalls

Frequently asked questions

Theory