Increasing, decreasing, stationary points Worked Solutions — A-Level Maths

Fully worked, step-by-step solutions to A-Level Increasing, decreasing, stationary points questions. See exactly how to solve problems on quadratic, turning point, minimum value, maximum value.

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A-Level70 questionsStep-by-step solutions

Question 1

3 markseasy

The curve

C

has equation

y=x^{2}-4x+7

. Find the coordinates of the stationary point of

C

Worked solution

Differentiate the function
$\frac{dy}{dx}=2 x - 4$
A stationary point is where the gradient is zero, so first we need the gradient function $\frac{dy}{dx}$ . Recall the rule: multiply by the power, then subtract one from the power.
Set the gradient equal to zero
$2 \left(x - 2\right)=0$
At a stationary point the tangent is flat, so its gradient is $0$ . We set the derivative to zero and solve to find where those points are.
Solve for $x$
$x=2$
Rearranging gives the single $x$ -coordinate where the curve is stationary.
Find the $y$ -coordinate(s)
$y=3\ \text{at }x=2$
Substitute each $x$ back into the original curve to get the full coordinates. Always use the original $y$ , not the derivative.
Find the second derivative
$\frac{d^2y}{dx^2}=2$
To decide whether each point is a maximum or minimum we use the second derivative test, so we differentiate the gradient function once more.
Apply the second derivative test
$x=2:\ 2>0\Rightarrow\text{minimum}$
If $\frac{d^2y}{dx^2}>0$ the point is a minimum (curve bends upwards); if it is $<0$ it is a maximum. Substitute each $x$ into the second derivative and read off the sign.

Answer

The minimum point is at (2,\ 3).

Question 2

3 markseasy

Find the coordinates of the turning point of the curve

y=x^{2}-6x+5

Worked solution

Differentiate the function
$\frac{dy}{dx}=2 x - 6$
A stationary point is where the gradient is zero, so first we need the gradient function $\frac{dy}{dx}$ . Recall the rule: multiply by the power, then subtract one from the power.
Set the gradient equal to zero
$2 \left(x - 3\right)=0$
At a stationary point the tangent is flat, so its gradient is $0$ . We set the derivative to zero and solve to find where those points are.
Solve for $x$
$x=3$
Rearranging gives the single $x$ -coordinate where the curve is stationary.
Find the $y$ -coordinate(s)
$y=-4\ \text{at }x=3$
Substitute each $x$ back into the original curve to get the full coordinates. Always use the original $y$ , not the derivative.
Find the second derivative
$\frac{d^2y}{dx^2}=2$
To decide whether each point is a maximum or minimum we use the second derivative test, so we differentiate the gradient function once more.
Apply the second derivative test
$x=3:\ 2>0\Rightarrow\text{minimum}$
If $\frac{d^2y}{dx^2}>0$ the point is a minimum (curve bends upwards); if it is $<0$ it is a maximum. Substitute each $x$ into the second derivative and read off the sign.

Answer

The minimum point is at (3,\ -4).

Question 3

3 markseasy

Find the minimum value of

y=x^{2}+8x+3

Worked solution

Differentiate the function
$\frac{dy}{dx}=2 x + 8$
A stationary point is where the gradient is zero, so first we need the gradient function $\frac{dy}{dx}$ . Recall the rule: multiply by the power, then subtract one from the power.
Set the gradient equal to zero
$2 \left(x + 4\right)=0$
At a stationary point the tangent is flat, so its gradient is $0$ . We set the derivative to zero and solve to find where those points are.
Solve for $x$
$x=-4$
Rearranging gives the single $x$ -coordinate where the curve is stationary.
Find the $y$ -coordinate(s)
$y=-13\ \text{at }x=-4$
Substitute each $x$ back into the original curve to get the full coordinates. Always use the original $y$ , not the derivative.
Find the second derivative
$\frac{d^2y}{dx^2}=2$
To decide whether each point is a maximum or minimum we use the second derivative test, so we differentiate the gradient function once more.
Apply the second derivative test
$x=-4:\ 2>0\Rightarrow\text{minimum}$
If $\frac{d^2y}{dx^2}>0$ the point is a minimum (curve bends upwards); if it is $<0$ it is a maximum. Substitute each $x$ into the second derivative and read off the sign.

Answer

y=-13

Question 4

3 markseasy

The curve has equation

y=2x^{2}-12x+7

. Find the coordinates of its stationary point.

Worked solution

Differentiate the function
$\frac{dy}{dx}=4 x - 12$
A stationary point is where the gradient is zero, so first we need the gradient function $\frac{dy}{dx}$ . Recall the rule: multiply by the power, then subtract one from the power.
Set the gradient equal to zero
$4 \left(x - 3\right)=0$
At a stationary point the tangent is flat, so its gradient is $0$ . We set the derivative to zero and solve to find where those points are.
Solve for $x$
$x=3$
Rearranging gives the single $x$ -coordinate where the curve is stationary.
Find the $y$ -coordinate(s)
$y=-11\ \text{at }x=3$
Substitute each $x$ back into the original curve to get the full coordinates. Always use the original $y$ , not the derivative.
Find the second derivative
$\frac{d^2y}{dx^2}=4$
To decide whether each point is a maximum or minimum we use the second derivative test, so we differentiate the gradient function once more.
Apply the second derivative test
$x=3:\ 4>0\Rightarrow\text{minimum}$
If $\frac{d^2y}{dx^2}>0$ the point is a minimum (curve bends upwards); if it is $<0$ it is a maximum. Substitute each $x$ into the second derivative and read off the sign.

Answer

The minimum point is at (3,\ -11).

Question 5

3 markseasy

Find the maximum value of

y=-x^{2}+10x-3

Worked solution

Differentiate the function
$\frac{dy}{dx}=10 - 2 x$
A stationary point is where the gradient is zero, so first we need the gradient function $\frac{dy}{dx}$ . Recall the rule: multiply by the power, then subtract one from the power.
Set the gradient equal to zero
$2 \left(5 - x\right)=0$
At a stationary point the tangent is flat, so its gradient is $0$ . We set the derivative to zero and solve to find where those points are.
Solve for $x$
$x=5$
Rearranging gives the single $x$ -coordinate where the curve is stationary.
Find the $y$ -coordinate(s)
$y=22\ \text{at }x=5$
Substitute each $x$ back into the original curve to get the full coordinates. Always use the original $y$ , not the derivative.
Find the second derivative
$\frac{d^2y}{dx^2}=-2$
To decide whether each point is a maximum or minimum we use the second derivative test, so we differentiate the gradient function once more.
Apply the second derivative test
$x=5:\ -2<0\Rightarrow\text{maximum}$
If $\frac{d^2y}{dx^2}>0$ the point is a minimum (curve bends upwards); if it is $<0$ it is a maximum. Substitute each $x$ into the second derivative and read off the sign.

Answer

y=22

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