New Zealand
Level 7 - NCEA Level 2

# Tangents (gradients and equations using power rule)

Lesson

This means using the derivative we can

• find the gradient function, also called the derivative
• evaluate the gradient at a point, say $x=a$x=a
• find the equation of the tangent at that point, because we have the gradient and a point

Let's work through an example

Find the equation of the tangent to the curve $f(x)=\frac{1}{4}x^2-2x+3$f(x)=14x22x+3 at the point where $x=4$x=4

Think:

This is our plan of attack

a) find the derivative function $f'(x)$f(x)

b) find the value of the derivative at the point where $x=4$x=4, this will be the gradient of the tangent

c) find the full coordinate on the curve by finding $f(4)$f(4)

d) use the point and the gradient to find the equation of the tangent.

a) find the derivative function $f'(x)$f(x)

The function is $f(x)=\frac{1}{4}x^2-2x+3$f(x)=14x22x+3, so using the power rule on each term we find that the derivative is $f'(x)=\frac{x}{2}-2$f(x)=x22

b) find the value of the derivative at the point where $x=4$x=4, this will be the gradient of the tangent.

$f'(x)=\frac{x}{2}-2$f(x)=x22, so $f'(4)=\frac{4}{2}-2=0$f(4)=422=0

This means that at the point where x=4, the gradient is 0, meaning that the tangent at that point is horizontal.

c) find the full coordinate on the curve by finding f(4)

The function is $f(x)=\frac{1}{4}x^2-2x+3$f(x)=14x22x+3, so $f(4)=\frac{1}{4}4^2-2\times4+3=4-8+3=-1$f(4)=14422×4+3=48+3=1.  So the coordinate on the curve is $(4,-1)$(4,1)

d) use the point and the gradient to find the equation of the tangent.

The point is $(4,-1)$(4,1) and the gradient is $0$0.  The equation of the horizontal line is $y=-1$y=1

#### Worked Examples

##### Question 1

By considering the graph of $f\left(x\right)=-6$f(x)=6, find $f'\left(4\right)$f(4).

##### Question 2

By considering the graph of $f\left(x\right)=2x-3$f(x)=2x3, find $f'$f$\left(-4\right)$(4).

##### Question 3

Consider the parabola $f\left(x\right)=x^2+3x-10$f(x)=x2+3x10.

1. Solve for the $x$x-intercepts. Write all solutions on the same line, separated by a comma.

2. Determine the gradient of the tangent at the positive $x$x-intercept.

##### Question 4

Consider the tangent to the curve $f\left(x\right)=5\sqrt{x}$f(x)=5x at the point $\left(\frac{1}{9},\frac{5}{3}\right)$(19,53).

1. Firstly, find the gradient of the function $f\left(x\right)=5\sqrt{x}$f(x)=5x at $x=\frac{1}{9}$x=19.

2. Hence find the equation of the tangent to the curve $f\left(x\right)=5\sqrt{x}$f(x)=5x at the point $\left(\frac{1}{9},\frac{5}{3}\right)$(19,53).

Express the equation of the tangent line in the form $y=mx+c$y=mx+c.

##### Question 5

Consider the tangent to the curve $f\left(x\right)=\frac{2}{x^3}$f(x)=2x3 at $x=-2$x=2.

1. Firstly, find the gradient of the function $f\left(x\right)=\frac{2}{x^3}$f(x)=2x3 at $x=-2$x=2.

2. Determine the $y$y-coordinate of the point on the tangent line.

3. Hence find the equation of the tangent to the curve $f\left(x\right)=\frac{2}{x^3}$f(x)=2x3 at $x=-2$x=2.

Express the equation of the tangent line in the form $y=mx+c$y=mx+c.

### Outcomes

#### M7-10

Apply differentiation and anti-differentiation techniques to polynomials

#### 91262

Apply calculus methods in solving problems