Slopes & Tangents — Mathslive.ie

Section 1 of 8

Slope of a Curve

A straight line has the same slope everywhere. A curve doesn't — its slope changes as you move along it.

To get the slope of a curve at a specific point, we differentiate.

Slope of a curve

$m = \dfrac{dy}{dx}$

Differentiate $y$, then sub in the $x$-value of the point.

A few facts about slope you already know:

$m = \dfrac{y_{2} - y_{1}}{x_{2} - x_{1}} \;$ (slope from two points)

$y = mx + c \;$ (equation of a line, slope $m$)

$m = \tan\theta \;$ (slope = tan of angle with positive $x$-axis)

Section 2 of 8

Slope at a given $x$-value

Find the slope of $\;y = x^{2} + 3x + 1\;$ when (i) $x = -1$ (ii) $x = 1$.

$y = x^{2} + 3x + 1$

$\dfrac{dy}{dx} = 2x + 3$

Sub in each $x$-value:

$x = -1: \;\; m = 2(-1) + 3 = 1$

$x = 1: \;\; m = 2(1) + 3 = 5$

So the curve is steeper at $x=1$ than at $x=-1$.

Find the slope to $\;y = x^{2} - 2x + 1\;$ when $x = -1$ and $x = 3$.

$\dfrac{dy}{dx} = 2x - 2$

$x = -1: \;\; m = 2(-1) - 2 = -4$

$x = 3: \;\; m = 2(3) - 2 = 4$

The curve is a U-shape with minimum at $x = 1$. At $x = -1$ the slope is negative (going down). At $x = 3$ the slope is positive (going up). Same size, opposite signs — the parabola is symmetric.

You try

Find the slope of $y = x^{2} + 5x - 2$ at $x = 2$.

Differentiate, then sub $x = 2$.

$\dfrac{dy}{dx} = 2x + 5$

$x = 2: \;\; m = 2(2) + 5$

$m = 9$

You try

Find the slope of $y = x^{3} - 4x + 1$ at $x = -2$.

$\dfrac{dy}{dx} = 3x^{2} - 4$. Be careful with the sign of $x^{2}$ when $x$ is negative.

$\dfrac{dy}{dx} = 3x^{2} - 4$

$x = -2: \;\; m = 3(-2)^{2} - 4 = 3(4) - 4$

$m = 8$

You try

Find the slope of $y = 2x^{2} - 7x + 3$ at $x = -1$ and at $x = 4$.

Differentiate once, then sub each value.

$\dfrac{dy}{dx} = 4x - 7$

$x = -1: \;\; m = 4(-1) - 7 = -11$

$x = 4: \;\; m = 4(4) - 7 = 9$

$m = -11$ and $m = 9$

Section 3 of 8

Finding a point with a given slope

Same idea, run backwards: we're told the slope; find where on the curve it happens.

Find the point on $\;y = x^{2} - 3x + 5\;$ where the slope is $7$.

$y = x^{2} - 3x + 5$

$\dfrac{dy}{dx} = 2x - 3$

Set the slope equal to $7$ and solve for $x$:

$2x - 3 = 7$

$x = 5$

Now find $y$ — sub $x = 5$ back into the original curve (not the derivative):

$y = (5)^{2} - 3(5) + 5 = 15$

Point: $(5, 15)$

You try

Find the point on $y = x^{2} - 5x + 2$ where the slope is $3$.

Set $\dfrac{dy}{dx} = 3$, solve for $x$, then sub back to find $y$.

$\dfrac{dy}{dx} = 2x - 5$

$2x - 5 = 3$

$x = 4$

$y = (4)^{2} - 5(4) + 2 = -2$

$(4, -2)$

You try

Find the point on $y = x^{3} - 6x + 1$ where the slope is $6$, taking the positive value of $x$.

A cubic gives a quadratic when differentiated — expect two $x$-values.

$\dfrac{dy}{dx} = 3x^{2} - 6$

$3x^{2} - 6 = 6$

$3x^{2} = 12$

$x^{2} = 4 \;\;\Rightarrow\;\; x = \pm 2$

Take $x = 2$: $\;\; y = (2)^{3} - 6(2) + 1 = -3$

$(2, -3)$

Section 4 of 8

Tangent parallel to a given line

Parallel lines have the same slope. So if the tangent is parallel to a given line, the curve's slope equals the line's slope at that point.

Find the point on $\;y = x^{2} - 6x + 1\;$ where the tangent is parallel to $\;2x - y = 9$.

First find the slope of the line. Rearrange to $y = mx + c$ form:

$2x - y = 9$

$-y = -2x + 9$

$y = 2x - 9$

Line slope $m = 2$

Now match the curve's slope to that:

$\dfrac{dy}{dx} = 2x - 6$

$2x - 6 = 2$

$2x = 8 \;\;\Rightarrow\;\; x = 4$

$y = (4)^{2} - 6(4) + 1 = -7$

Point: $(4, -7)$

You try

Find the point on $y = x^{2} + 4x - 3$ where the tangent is parallel to $\;y = 6x + 1$.

Line is already in $y = mx + c$ form — read off the slope.

Line slope: $m = 6$

$\dfrac{dy}{dx} = 2x + 4 = 6$

$x = 1$

$y = (1)^{2} + 4(1) - 3 = 2$

$(1, 2)$

You try

Find the point on $y = x^{2} - 8x + 5$ where the tangent is parallel to $\;4x + y = 1$.

Watch the sign — rearrange the line to $y = mx + c$ first.

$4x + y = 1 \;\;\Rightarrow\;\; y = -4x + 1$

Line slope: $m = -4$

$\dfrac{dy}{dx} = 2x - 8 = -4$

$2x = 4 \;\;\Rightarrow\;\; x = 2$

$y = (2)^{2} - 8(2) + 5 = -7$

$(2, -7)$

Section 5 of 8

What is a tangent?

A tangent is a line that touches a curve at one point.

To write the equation of a tangent you need

$\bullet$ the slope (from $\dfrac{dy}{dx}$ at the given $x$)

$\bullet$ a point (the $(x, y)$ where it touches)

Then use the point–slope form: $\;y - y_{1} = m(x - x_{1})$.

Section 6 of 8

Equation of the tangent

Find the tangent to $\;y = x^{2} - 5x + 1\;$ when $x = -2$.

Get the slope:

$\dfrac{dy}{dx} = 2x - 5$

$x = -2: \;\; m = 2(-2) - 5 = -9$

Get the point — sub $x = -2$ into the original:

$y = (-2)^{2} - 5(-2) + 1 = 4 + 10 + 1 = 15$

Point: $(-2, 15)$

Tangent equation $y - y_{1} = m(x - x_{1})$:

$y - 15 = -9(x + 2)$

Find the tangent to $\;y = x^{3} - 2x^{2} + 5\;$ when $x = -1$.

$\dfrac{dy}{dx} = 3x^{2} - 4x$

$x = -1: \;\; m = 3(-1)^{2} - 4(-1) = 3 + 4 = 7$

$y = (-1)^{3} - 2(-1)^{2} + 5 = -1 - 2 + 5 = 2$

Point: $(-1, 2)$

$y - 2 = 7(x + 1)$

$y - 2 = 7x + 7$

$y = 7x + 9$

You try

Find the equation of the tangent to $y = x^{2} + 2x - 3$ at $x = 1$.

You need slope and point. Then $y - y_{1} = m(x - x_{1})$.

$\dfrac{dy}{dx} = 2x + 2$

$x = 1: \;\; m = 4$

$y = (1)^{2} + 2(1) - 3 = 0$

Point: $(1, 0)$

$y - 0 = 4(x - 1)$

$y = 4x - 4$

You try

Find the equation of the tangent to $y = 2x^{2} - 3x + 1$ at $x = 2$.

Same routine: slope from $\dfrac{dy}{dx}$, then point from the original.

$\dfrac{dy}{dx} = 4x - 3$

$x = 2: \;\; m = 5$

$y = 2(4) - 3(2) + 1 = 3$

Point: $(2, 3)$

$y - 3 = 5(x - 2)$

$y = 5x - 7$

You try

Find the equation of the tangent to $y = x^{3} - 4x + 6$ at $x = 2$.

$\dfrac{dy}{dx} = 3x^{2} - 4$. Then sub.

$\dfrac{dy}{dx} = 3x^{2} - 4$

$x = 2: \;\; m = 3(4) - 4 = 8$

$y = (2)^{3} - 4(2) + 6 = 6$

Point: $(2, 6)$

$y - 6 = 8(x - 2)$

$y = 8x - 10$

Section 7 of 8

Angle with the positive $x$-axis

Every line has an angle it makes with the positive $x$-axis. The link to slope is:

Slope as an angle

$m = \tan\theta$

$\theta$ measured from the positive $x$-axis, going anticlockwise.

Two cases to remember:

Line going up to the right: $\;\;0° < \theta < 90° \;\;\Rightarrow\;\; \tan\theta > 0 \;\;\Rightarrow\;\; m > 0$

Line going down to the right: $\;\;90° < \theta < 180° \;\;\Rightarrow\;\; \tan\theta < 0 \;\;\Rightarrow\;\; m < 0$

Find the point on $\;y = x^{2} - 3x + 1\;$ where the tangent forms an angle of $135°$ with the positive $x$-axis.

Turn the angle into a slope:

$m = \tan 135° = -1$

Now match the curve's slope:

$\dfrac{dy}{dx} = 2x - 3$

$2x - 3 = -1$

$x = 1$

$y = (1)^{2} - 3(1) + 1 = -1$

Point: $(1, -1)$

You try

Find the point on $y = x^{2} + 5x + 2$ where the tangent forms an angle of $45°$ with the positive $x$-axis.

$\tan 45° = 1$. So you're finding where the slope is $1$.

$m = \tan 45° = 1$

$\dfrac{dy}{dx} = 2x + 5 = 1$

$x = -2$

$y = (-2)^{2} + 5(-2) + 2 = -4$

$(-2, -4)$

You try

Find the point on $y = x^{2} + x - 4$ where the tangent forms an angle of $135°$ with the positive $x$-axis.

$\tan 135° = -1$.

$m = \tan 135° = -1$

$\dfrac{dy}{dx} = 2x + 1 = -1$

$x = -1$

$y = (-1)^{2} + (-1) - 4 = -4$

$(-1, -4)$

You try

Find the equation of the tangent to $y = x^{2} - 4x + 5$ at the point where the tangent forms an angle of $45°$ with the positive $x$-axis.

Find $x$ where slope is $1$, get the point, then write the tangent line.

$m = \tan 45° = 1$

$\dfrac{dy}{dx} = 2x - 4 = 1$

$x = \dfrac{5}{2}$

$y = \left(\dfrac{5}{2}\right)^{2} - 4\left(\dfrac{5}{2}\right) + 5 = \dfrac{25}{4} - 5 = \dfrac{5}{4}$

Point: $\left(\dfrac{5}{2}, \dfrac{5}{4}\right)$

$y - \dfrac{5}{4} = 1\left(x - \dfrac{5}{2}\right)$

$y = x - \dfrac{5}{4}$

Section 8 of 8

Wrap-up

Every problem in this lesson came down to two ingredients: a slope (from $\dfrac{dy}{dx}$) and a point (from the original curve). Whether the question gives you the $x$-value, the slope, a parallel line, or an angle — you're always finding one of those two things first, then the other.

Quick reference

$\bullet$ Slope at a point: differentiate, sub $x$ in.

$\bullet$ Point with given slope: set $\dfrac{dy}{dx} = m$, solve, then sub $x$ into original.

$\bullet$ Tangent parallel to a line: line's slope = curve's slope.

$\bullet$ Tangent at angle $\theta$: slope $= \tan\theta$.

$\bullet$ Tangent equation: $y - y_{1} = m(x - x_{1})$.

That's Slopes & Tangents.

Next lesson builds on this: when slope is zero you get a turning point — minimum or maximum. We'll also look at where curves are increasing and decreasing.