2.14.5 Tangents
Tangent:
A tangent to a curve is a straight line that touches the curve only once without crossing or intersecting it.
Gradient:
The gradient is a vector that represents a function’s rate and direction of change.
- For finding the gradient of a non-linear graph, we use a tangent to its curve.
- Gradient of a curve at a point equals the gradient of the tangent at that point.
To find gradient of a nonlinear graph at a point
- Draw a tangent to the curve at the given point.
- Find the gradient of the tangent using
\( gradient= \frac{rise}{run} \)
- Put sign of gradient depending upon the slope of tangent
Worked example:
The diagram shows the graph of a function.
By drawing a suitable tangent, find an estimate for the gradient of the function at the point P.
Draw the tangent at point P.
Take any two points on the tangent line and find the gradient by using \( gradient= \frac{y_2-y_1}{x_2-x_1} \)
\( gradient= \frac{1.1-(-0.4)}{4.8-2} \)
\( = \frac{(1.1+0.4)}{4.8-2} \)
\( = \frac{(1.5)}{2.8} \)
\(=0.54\)
The gradient is \(0.54\)
Test yourself
Question 1:
By drawing a suitable tangent, estimate the gradient of the curve at the point B.
[3]
Question 2:
- Complete the table of values for \(y=\frac{x^3}{3}-\frac{1}{2x^2}, x≠0 .\)
[4]
- On the grid draw the graph of \(y=\frac{x^3}{3}-\frac{1}{2x^2} for -3≤x≤-0.3\ and\ 0.3≤x≤3 .\)
[5]
- By drawing a suitable tangent, find an estimate of the gradient of the curve at \(x=-2 .\)
[3]
- Write down the equation of the tangent to the curve at \(x=-2 .\)
Give your answer in the form in the form \(y=mx+c\) .
[2]
- Use your graph to solve the equations.
\(\frac{x^3}{3}-\frac{1}{2x^2}=0\)
[1]
\(\frac{x^3}{3}-\frac{1}{2x^2}+=0\)
[3]
Question 3:
The table shows some values for \(y=x^3+x^2-5x\).
Complete the table.
[3]
On the grid, draw the graph of \(y=x^3+x^2-5x\ for\ -3≤x≤3.\)
[4]
Use your graph to solve the equation \(y=x^3+x^2-5x=0\).
[2]
By drawing a suitable tangent, find an estimate of the gradient of the curve at \(x=2.\)
[3]
Question 4:
The table shows some values for \(y=\frac{3}{10}x^3-2x\ for\ -3≤x≤3\) .
Complete the table.
[4]
On the grid, draw the graph of \(y=\frac{3}{10}x^3-2x\ for\ -3≤x≤3\) .
[4]
On the grid, draw a suitable straight line to solve the equation
\(\frac{3}{10}x^3-2x=\frac{1}{2}(1-x)\ for\ -3≤x≤3\) .
[4]
\(for -3≤x≤3\), the equation \(\frac{3}{10}x^3-2x=1\) has n solutions.
Write down the value of n.
[1]
Question 5:
The table shows some values for \(y=2x+\frac{1}{x}-3\ for\ 0.125≤x≤3\) .
Complete the table.
[3]
On the grid, draw the graph of \(y=2x+\frac{1}{x}-3\ for\ 0.125≤x≤3\) .
[4]
Use your graph to solve \(y=2x+\frac{1}{x}–3≥2\) .
[3]
The equation \(\frac{1}{x}=7-3x\) can be solved using your graph in part band a straight line .
- Write down the equation of this straight line.
[2]
- Draw this straight line and solve the equation \(\frac{1}{x}=7-3x\).
[3]