Tangents (gradients and equations using power rule)
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)=14x2−2x+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)=14x2−2x+3, so using the power rule on each term we find that the derivative is $f'(x)=\frac{x}{2}-2$f′(x)=x2−2
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)=x2−2, so $f'(4)=\frac{4}{2}-2=0$f′(4)=42−2=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)=14x2−2x+3, so $f(4)=\frac{1}{4}4^2-2\times4+3=4-8+3=-1$f(4)=1442−2×4+3=4−8+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)=2x−3, find $f'$f′$\left(-4\right)$(−4).
Question 3
Consider the parabola $f\left(x\right)=x^2+3x-10$f(x)=x2+3x−10.
Solve for the $x$x-intercepts. Write all solutions on the same line, separated by a comma.
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)=5√x at the point $\left(\frac{1}{9},\frac{5}{3}\right)$(19,53).
Firstly, find the gradient of the function $f\left(x\right)=5\sqrt{x}$f(x)=5√x at $x=\frac{1}{9}$x=19.
Hence find the equation of the tangent to the curve $f\left(x\right)=5\sqrt{x}$f(x)=5√x 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.
Firstly, find the gradient of the function $f\left(x\right)=\frac{2}{x^3}$f(x)=2x3 at $x=-2$x=−2.
Determine the $y$y-coordinate of the point on the tangent line.
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.