A tangent to a function is a straight line and as such we can use our knowledge of linear functions to find the equation of a tangent. We can use differentiation to find the gradient of the tangent to a function at any given point.
For a function $y=f(x)$y=f(x) the equation of the tangent at the point of contact $(x_1,y_1)$(x1,y1) can be found using either:
Where the gradient of the tangent is $m=f'(x_1)$m=f′(x1).
From a graph look for two easily identifiable points then calculate the gradient using $m=\frac{rise}{run}$m=riserun. Then use the gradient and one of the points found in one of the forms above to find the equation. If it is clear on a graph a convenient point to use would be the $y$y-intercept.
Steps: The equation is of the form $y=mx+c$y=mx+c, we need to find $m$m and then $c$c.
Find the equation of the tangent to $f(x)$f(x) at the point pictured below.
Think: We can see the point of contact and the $y$y-intercept clearly. Use these two points to find the gradient and then write the equation in the form $y=mx+c$y=mx+c.
Do: The point of contact is $\left(1,-3\right)$(1,−3) and the $y$y-intercept is $\left(0,-5\right)$(0,−5). Thus, the gradient is:
$m$m | $=$= | $\frac{y_2-y_1}{x_2-x_1}$y2−y1x2−x1 |
$=$= | $\frac{-3-\left(-5\right)}{1-0}$−3−(−5)1−0 | |
$=$= | $2$2 |
Since we have the $y$y-intercept we know $c=-5$c=−5, and hence the equation of the tangent is $y=2x-5$y=2x−5.
Find the equation of the tangent to $f(x)=x^2+5x-3$f(x)=x2+5x−3 at the point $\left(2,11\right)$(2,11).
Think: We have been given the point of contact, $\left(2,11\right)$(2,11), we also require the gradient of the tangent. So we will find the derivative and evaluate it at $x=2$x=2. Then we can use both the gradient and point of contact to find the equation of the tangent.
Do:
$f'(x)$f′(x) | $=$= | $2x+5$2x+5 |
Thus, $m$m | $=$= | $f'(2)$f′(2) |
$=$= | $2(2)+5$2(2)+5 | |
$=$= | $9$9 |
We know the tangent has the form $y=9x+c$y=9x+c and goes through the point, $\left(2,11\right)$(2,11). Substituting the point of contact in we can find $c$c.
$11$11 | $=$= | $9\left(2\right)+c$9(2)+c |
$11$11 | $=$= | $18+c$18+c |
$\therefore c$∴c | $=$= | $-7$−7 |
Hence, the equation of the tangent to $f(x)$f(x) at $x=2$x=2, is $y=9x-7$y=9x−7.
Find the equation of the tangent to $f(x)=\sqrt{x}$f(x)=√x at $x=4$x=4.
Think: This time we have not been given the point of contact but we can evaluate the function at $x=4$x=4 to find it. We also need to find the derivative to find the gradient of the tangent.
Do:
Find the gradient of tangent:
$f(x)$f(x) | $=$= | $x^{\frac{1}{2}}$x12 |
$f'(x)$f′(x) | $=$= | $\frac{1}{2}x^{-\frac{1}{2}}$12x−12 |
$=$= | $\frac{1}{2\sqrt{x}}$12√x | |
$\therefore f'(4)$∴f′(4) | $=$= | $\frac{1}{2\sqrt{4}}$12√4 |
$=$= | $\frac{1}{4}$14 |
Find the point of contact, when $x=4$x=4:
$f(4)$f(4) | $=$= | $\sqrt{4}$√4 |
$=$= | $2$2 |
Thus, the point of contact is $(4,2)$(4,2).
Find $c$c:
The tangent is of the form $y=\frac{1}{4}x+c$y=14x+c and passes through $(4,2)$(4,2). Substituting into the equation we get:
$2$2 | $=$= | $\frac{1}{4}\left(4\right)+c$14(4)+c |
$2$2 | $=$= | $1+c$1+c |
$\therefore c$∴c | $=$= | $1$1 |
Hence, the equation of the tangent to $f(x)$f(x) at $x=4$x=4 is $y=\frac{1}{4}x+1$y=14x+1.
Consider the curve $f\left(x\right)$f(x) drawn below along with $g\left(x\right)$g(x), which is a tangent to the curve.
What are the coordinates of the point at which $g\left(x\right)$g(x) is a tangent to the curve $f\left(x\right)$f(x)?
Note that this point has integer coordinates. Give your answer in the form $\left(a,b\right)$(a,b).
What is the gradient of the tangent line?
Hence determine the equation of the line $y=g\left(x\right)$y=g(x).
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.
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.
Consider the curve $f\left(x\right)=x^2+8x+15$f(x)=x2+8x+15.
Find $f'\left(x\right)$f′(x).
Find the gradient of the tangent to the curve at the point $\left(4,63\right)$(4,63).
Determine the equation of the tangent to the curve $f\left(x\right)=x^2+8x+15$f(x)=x2+8x+15 at $\left(4,63\right)$(4,63).
Express the equation of the tangent line in the form $y=mx+c$y=mx+c.
Find the gradient of the normal to the curve at the point $\left(4,63\right)$(4,63).
Determine the equation of the normal to the curve $f\left(x\right)=x^2+8x+15$f(x)=x2+8x+15 at $\left(4,63\right)$(4,63).
Common problem solving questions include using the derivative to determine the point(s) on a curve where a given gradient occurs or using the information about gradients and the original function to determine unknown coefficients in the original function.
The function $f\left(x\right)=x^3+ax^2+bx+c$f(x)=x3+ax2+bx+c has a $y$y-intercept of $3$3. The function has gradient of zero at $x=1$x=1 and a root at $x=-3$x=−3. Determine the values of $a$a, $b$band $c$c.
Think: Break the information into parts and determine if the information given is about the function itself or its derivative.
We have three pieces of information and three unknowns, so we should be able to solve using simultaneous equations.
Do: We can begin by using the information about the $y$y-intercept. Substituting $\left(0,3\right)$(0,3) into the original function we get:
$0^3+a\times0^2+b\times0+c$03+a×02+b×0+c | $=$= | $3$3 |
$\therefore c$∴c | $=$= | $3$3 |
To use the information about the gradient, we will need to first find ourselves the derivative.
$f'\left(x\right)=3x^2+2ax+b$f′(x)=3x2+2ax+b
Using the information $f'(1)=0$f′(1)=0, we obtain the equation:
$3(1)^2+2a(1)+b$3(1)2+2a(1)+b | $=$= | $0$0 | |
$2a+b$2a+b | $=$= | $-3$−3 | ....Equation $1$1 |
To use the information about the $x$x-intercept, we can substitute $\left(-3,0\right)$(−3,0) into the original function to obtain the equation:
$\left(-3\right)^3+a\left(-3\right)^2+b\left(-3\right)+3$(−3)3+a(−3)2+b(−3)+3 | $=$= | $0$0 | |
$-27+9a-3b+3$−27+9a−3b+3 | $=$= | $0$0 | |
$9a-3b$9a−3b | $=$= | $24$24 | ....Equation $2$2 |
Solving equation $1$1 and $2$2 simultaneously (either with the elimination method, substitution method or with technology) we find that $a=1$a=1 and $b=-5$b=−5. Thus the original function was $f(x)=x^3+x^2-5x+3$f(x)=x3+x2−5x+3.
Consider the function $f\left(x\right)=x^2+5x$f(x)=x2+5x.
Find the $x$x-coordinate of the point at which $f\left(x\right)$f(x) has a gradient of $13$13.
Hence state the coordinates of the point on the curve where the gradient is $13$13.
The curve $y=ax^3+bx^2+2x-17$y=ax3+bx2+2x−17 has a gradient of $58$58 at the point $\left(2,31\right)$(2,31).
Use the fact that the gradient of the curve at the point $\left(2,31\right)$(2,31) is $58$58 to express $b$b in terms of $a$a.
Use the fact that the curve passes through the point $\left(2,31\right)$(2,31) to express $b$b in terms of $a$a.
Hence solve for $a$a.
Hence solve for $b$b.