Quadratic Functions Worked Examples (KBAT)

Q: How do I show a line is a tangent to a curve?

Solve the line and curve simultaneously to get one quadratic, then set its discriminant b2−4ac=0b^2-4ac=0b2−4ac=0. Equal roots mean the line meets the curve at exactly one point, which is tangency.

Q: Why factor out the leading coefficient before completing the square?

Completing the square needs the x2x^2x2 coefficient to be 111 inside the bracket. Factoring out −1-1−1 (or any aaa) first keeps the algebra correct; the common slip is forgetting to expand it back over both terms.

Q: When does \alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta help?

Whenever a question gives a symmetric expression in the roots. It lets you answer using only the sum and product from the coefficients, so you never need the roots themselves.

Q: Should I give both values when k^2=25?

Yes, unless the question restricts the sign. Here k=5k=5k=5 and k=−5k=-5k=−5 both satisfy α2+β2=13\alpha^2+\beta^2=13α2+β2=13, so both are valid answers and each carries marks.

How to use this set

These three hard Quadratic Functions problems each fold two ideas together. The first uses simultaneous equations and the discriminant to make a line touch a curve; the second combines completing the square with a stated maximum value; the third links the sum and product of roots to the symmetric expression $\alpha^2+\beta^2$ .

They are demanding but every line is shown, so nothing is left as a leap.

Attempt each on paper first, set up your own equations, then compare with ours. Hard questions reward a clear plan: name what condition the situation forces (tangency means equal roots, a maximum means a negative leading coefficient), write it as an equation, and solve.

Checking your answer by substitution at the end is what separates a full-mark solution from a near miss.

Three worked examples

Q1[4 marks]

The straight line $y = 2x + k$ is a tangent to the curve $y = x^2 - 2x + 6$ . Find the value of $k$ .

Show worked solution

Where line and curve meet, their $y$ -values are equal. Set the expressions equal and collect everything on one side:

x^2 - 2x + 6 = 2x + k

x^2 - 2x - 2x + 6 - k = 0 \;\Rightarrow\; x^2 - 4x + (6-k) = 0

A tangent touches the curve at exactly one point, so this quadratic must have equal roots. That means the discriminant is zero.

Here $a=1$ , $b=-4$ , $c=6-k$ :

b^2 - 4ac = (-4)^2 - 4(1)(6-k) = 0

16 - 24 + 4k = 0 \;\Rightarrow\; 4k - 8 = 0 \;\Rightarrow\; k = 2

Check: with $k=2$ the quadratic is $x^2-4x+4=0$ , i.e. $(x-2)^2=0$ , giving the single point $x=2$ . There the line gives $y=2(2)+2=6$ and the curve gives $y=4-4+6=6$ , they touch at $(2,6)$ , confirming tangency.

Q2[4 marks]

The quadratic function $f(x) = -x^2 + 6x + c$ has a maximum value of $13$ . Find the value of $c$ and state the coordinates of the maximum point.

Show worked solution

Because the coefficient of $x^2$ is negative, the graph opens downward and has a maximum. Complete the square.

Factor $-1$ from the $x$ -terms:

f(x) = -\left(x^2 - 6x\right) + c

Inside the bracket, $x^2-6x = (x-3)^2 - 9$ . Substitute and expand the $-1$ carefully:

f(x) = -\left[(x-3)^2 - 9\right] + c = -(x-3)^2 + 9 + c

The term $-(x-3)^2$ is at most $0$ , reached when $x=3$ , so the maximum value of $f(x)$ is $9+c$ . Set this equal to the given maximum $13$ :

9 + c = 13 \;\Rightarrow\; c = 4

The maximum point is at $x=3$ , value $13$ , so it is $(3,\,13)$ . Check: $f(3) = -(3)^2 + 6(3) + 4 = -9 + 18 + 4 = 13$ , as required.

Q3[4 marks]

The roots of $x^2 - kx + 6 = 0$ are $\alpha$ and $\beta$ . Given that $\alpha^2 + \beta^2 = 13$ , find the possible values of $k$ .

Show worked solution

Read off the sum and product of roots. With $a=1$ , $b=-k$ , $c=6$ :

\alpha+\beta = -\frac{-k}{1} = k, \qquad \alpha\beta = \frac{6}{1} = 6

Express $\alpha^2+\beta^2$ through the sum and product using the identity $\alpha^2+\beta^2 = (\alpha+\beta)^2 - 2\alpha\beta$ :

\alpha^2 + \beta^2 = k^2 - 2(6) = k^2 - 12

Set this equal to the given value $13$ and solve:

k^2 - 12 = 13 \;\Rightarrow\; k^2 = 25 \;\Rightarrow\; k = \pm 5

Check $k=5$ : $x^2-5x+6=(x-2)(x-3)$ , roots $2$ and $3$ , and $2^2+3^2=4+9=13$ . Check $k=-5$ : $x^2+5x+6=(x+2)(x+3)$ , roots $-2$ and $-3$ , and $(-2)^2+(-3)^2=13$ .

Both values work.

Q4[4 marks]

A rectangular vegetable plot has length $x+8$ m and width $x$ m. The area of the plot must be less than $65\text{ m}^2$ .

Form a quadratic inequality in $x$ , and find the range of values of $x$ .

Show worked solution

Write the area as length times width, then form an inequality using the given upper bound. This gives a quadratic inequality in $x$ :

x(x+8) < 65

Expand and rearrange so one side is zero, then factorise the quadratic to find the critical values of $x$ :

x^2 + 8x - 65 < 0 \;\Rightarrow\; (x+13)(x-5) < 0

The expression is negative between its roots $-13$ and $5$ , so $-13<x<5$ ; since $x$ is a length it must also be positive, giving the combined range $0<x<5$ .

Answer

The width must satisfy $0 < x < 5$ . Sense check: at $x=4$ the plot is $4\text{ m}$ by $12\text{ m}$ , giving an area of $48\text{ m}^2 < 65\text{ m}^2$ , inside the range; at $x=5$ the area is exactly $65\text{ m}^2$ , so $x=5$ is correctly excluded.

Q5[5 marks]

The quadratic equation $px^2 - 4x + p = 0$ , where $p$ is a constant, has two distinct real roots. Find the range of values of $p$ .

Show worked solution

For the equation to have two distinct real roots, its discriminant must be positive, and the coefficient of $x^2$ must not be zero (otherwise the equation is linear, with at most one root). Here $a=p$ , $b=-4$ , $c=p$ :

b^2-4ac>0 \;\Rightarrow\; (-4)^2 - 4(p)(p) > 0

Simplify and solve this inequality for $p$ :

16 - 4p^2 > 0 \;\Rightarrow\; p^2 < 4 \;\Rightarrow\; -2 < p < 2

This range still includes $p=0$ , which must be excluded since the equation would then reduce to the linear equation $-4x=0$ with only one root, giving the final range $-2<p<0$ or $0<p<2$ .

Answer

$p$ satisfies $-2<p<0$ or $0<p<2$ . Sense check: $p=1$ gives $x^2-4x+1=0$ with discriminant $12>0$ , two distinct real roots; $p=3$ (outside the range) gives discriminant $16-36=-20<0$ , no real roots, as expected.

Q6[4 marks]

The roots of the quadratic equation $x^2 - 6x + 4 = 0$ are $\alpha$ and $\beta$ . Without solving for $\alpha$ and $\beta$ , form a quadratic equation whose roots are $\dfrac{1}{\alpha}$ and $\dfrac{1}{\beta}$ .

Show worked solution

Read the sum and product of roots from the given equation, then use $\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}$ and $\frac{1}{\alpha}\cdot\frac{1}{\beta}=\frac{1}{\alpha\beta}$ to find the sum and product of the new roots directly, without solving for $\alpha$ and $\beta$ themselves. With $a=1$ , $b=-6$ , $c=4$ :

\alpha+\beta = 6, \qquad \alpha\beta = 4

Substitute these into the reciprocal-sum and reciprocal-product formulas, then form the new equation using $x^2-(\text{sum})x+(\text{product})=0$ and multiply through by $4$ to clear fractions:

\frac{1}{\alpha}+\frac{1}{\beta} = \frac{\alpha+\beta}{\alpha\beta} = \frac{6}{4} = \frac{3}{2}, \qquad \frac{1}{\alpha}\cdot\frac{1}{\beta} = \frac{1}{\alpha\beta} = \frac{1}{4}

x^2 - \frac{3}{2}x + \frac{1}{4} = 0 \;\Rightarrow\; 4x^2 - 6x + 1 = 0

Answer

The required equation is $4x^2-6x+1=0$ . Sense check: substituting $x=\frac{1}{y}$ into the original equation and multiplying through by $y^2$ gives $4y^2-6y+1=0$ , the same equation, confirming the reciprocal-root transformation is correct.

Q7[6 marks]

A farmer has $40$ m of fencing to enclose a rectangular pen. One side of the pen lies along an existing straight wall, so fencing is needed only for the other three sides.

Let $x$ m be the length of each of the two sides perpendicular to the wall. Express the area of the pen, $A$ , in terms of $x$ , and hence find the maximum possible area.

Show worked solution

Let the side parallel to the wall have length $y$ m. Only three sides need fencing, so $2x+y=40$ , giving $y=40-2x$ .

The area is $A=xy$ :

y = 40 - 2x, \qquad A(x) = x(40-2x) = 40x - 2x^2

The coefficient of $x^2$ is negative, so $A(x)$ has a maximum. Factor out $-2$ and complete the square:

A(x) = -2x^2 + 40x = -2\left(x^2-20x\right) = -2\left[(x-10)^2 - 100\right] = -2(x-10)^2 + 200

Since $-2(x-10)^2 \le 0$ for every $x$ , the area is largest when $x=10$ , where $-2(x-10)^2=0$ . Then $y=40-2(10)=20$ .

Answer

The maximum area is $200\text{ m}^2$ , when $x=10$ m and $y=20$ m. Sense check: at $x=8$ , $y=24$ and the area is $192\text{ m}^2$ ; at $x=12$ , $y=16$ and the area is again $192\text{ m}^2$ , both less than $200\text{ m}^2$ , consistent with $x=10$ giving the maximum.

Q8[5 marks]

The curves $y = x^2 - 2x + 5$ and $y = -x^2 + 6x - 1$ intersect at two points. Find the coordinates of both points of intersection.

Show worked solution

At a point of intersection, both curves share the same $x$ and $y$ , so set the two expressions for $y$ equal:

x^2 - 2x + 5 = -x^2 + 6x - 1

Rearrange so one side is zero, simplify, and factorise to find the $x$ -coordinates; then substitute each value back into either curve to find the matching $y$ -coordinates:

2x^2 - 8x + 6 = 0 \;\Rightarrow\; x^2 - 4x + 3 = 0 \;\Rightarrow\; (x-1)(x-3) = 0 \;\Rightarrow\; x = 1 \text{ or } x = 3

x=1: \; y = 1^2 - 2(1) + 5 = 4 \qquad\qquad x=3: \; y = 3^2 - 2(3) + 5 = 8

Answer

The curves intersect at $(1,4)$ and $(3,8)$ . Sense check: the second curve gives $y=-1^2+6(1)-1=4$ at $x=1$ and $y=-3^2+6(3)-1=8$ at $x=3$ , matching both points, confirming they lie on both curves.

Key method points

A line is tangent to a curve when solving them simultaneously gives a quadratic with equal roots, so set its discriminant to $0$ .
For a downward parabola (negative $x^2$ coefficient), completing the square gives the maximum value as the constant after the squared term.
Factor out the leading coefficient before completing the square, and expand it back carefully, this is where signs go wrong.
Use $\alpha^2+\beta^2 = (\alpha+\beta)^2 - 2\alpha\beta$ to turn a symmetric condition into an equation in the sum and product.
A squared unknown, like $k^2=25$ , usually gives two answers $k=\pm 5$ , state both unless the question rules one out.

Common slips to avoid

Hard questions punish rushed algebra. When you form the quadratic for a tangent, collect every term on one side before setting the discriminant to zero.

When completing the square with a negative leading coefficient, factor out the $-1$ first and expand it back over both terms. And when a squared unknown appears, such as $k^2=25$ , do not stop at the positive root, state both $k=5$ and $k=-5$ unless the question rules one out.

How a teacher helps

Hard questions are really about choosing the right condition. Our teachers help you name it fast: tangency forces equal roots, a maximum forces a negative leading coefficient, a symmetric expression invites the sum-and-product identity.

We rehearse the sign-sensitive steps, factoring out $-1$ , expanding a completed square, until they stop tripping you up, and we insist on the final substitution check that catches a lost solution like $k=-5$ . Lessons are in English, and because Add Math is marked analytically, we make sure each decision and line earns its marks.

Get 1-to-1 help.

Book a Trial Class

Frequently asked questions

How do I show a line is a tangent to a curve?

Solve the line and curve simultaneously to get one quadratic, then set its discriminant $b^2-4ac=0$ . Equal roots mean the line meets the curve at exactly one point, which is tangency.

Why factor out the leading coefficient before completing the square?

Completing the square needs the $x^2$ coefficient to be $1$ inside the bracket. Factoring out $-1$ (or any $a$ ) first keeps the algebra correct; the common slip is forgetting to expand it back over both terms.

When does

\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta

help?

Whenever a question gives a symmetric expression in the roots. It lets you answer using only the sum and product from the coefficients, so you never need the roots themselves.

Should I give both values when

k^2=25

Yes, unless the question restricts the sign. Here $k=5$ and $k=-5$ both satisfy $\alpha^2+\beta^2=13$ , so both are valid answers and each carries marks.

Source:SRC-DSKP-EN

Written by the spmaddmath.com.my editorial team.· Last updated 5 September 2026

Quadratic Functions, Worked Examples (KBAT)

How to use this set

Three worked examples

Key method points

How a teacher helps

Frequently asked questions

Ready to get started?