Tangents, Points & Unknowns

Section 1 of 2

Tangent to a Curve

A tangent is a straight line that touches the curve at exactly one point — there the line and curve share the same slope. To write its equation you need a point and a slope.

Tangent = line

Slope: $m = \dfrac{dy}{dx}$. Equation: $y - y_1 = m(x - x_1)$. The slope comes from $\dfrac{dy}{dx}$; the point is given, or found by subbing the $x$-value into $y$.

Worked example — tangent at a given point

Find the tangent to $y = x^2 - 6x - 5$ at the point $(1, -10)$.

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

At $x = 1$: $m = 2(1) - 6 = -4$

$y - (-10) = -4(x - 1)$

$y + 10 = -4x + 4$

$4x + y = -6$

Worked example — find the point yourself

Find the tangent to $y = x^2 + 3x - 9$ when $x = 1$.

$\dfrac{dy}{dx} = 2x + 3$; at $x = 1$: $m = 5$

$y = 1^2 + 3(1) - 9 = -5$, point $(1, -5)$

$y + 5 = 5(x - 1) = 5x - 5$

$5x - y = 10$

You try

Find the tangent to $y = x^2 - 10x - 3$ when $x = 2$.

Slope from $\tfrac{dy}{dx}=2x-10$ at $x=2$. Find $y$ at $x=2$ for the point. Then $y-y_1=m(x-x_1)$.

$\dfrac{dy}{dx} = 2x - 10$; at $x=2$: $m = -6$

$y = 2^2 - 10(2) - 3 = -19$, point $(2, -19)$

$y + 19 = -6(x - 2) = -6x + 12$

$6x + y = -7$

You try

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

Cubic — same method, bigger algebra. $\tfrac{dy}{dx}=3x^2-10x-2$.

At $x=-1$: $m = 3(-1)^2 - 10(-1) - 2 = 11$

$y = (-1)^3 - 5(-1)^2 - 2(-1) + 1 = -3$, point $(-1,-3)$

$y + 3 = 11(x + 1)$

Section 2 of 2

Finding Points and Unknowns

Sometimes you’re given the slope and asked where it happens (a point). Other times the curve has an unknown letter and the slope is given. Either way: set $\dfrac{dy}{dx}$ equal to what you’re told and solve.

Worked example — find the point

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

$\dfrac{dy}{dx} = 2x - 7 = 5$

$2x = 12 \Rightarrow x = 6$

$y = 6^2 - 7(6) + 3 = -3$

$(6, -3)$

You try

Find the point on $y = x^2 - 11x + 3$ where the slope is $-7$.

Set $2x - 11 = -7$, solve for $x$, then sub back for $y$.

$2x - 11 = -7 \Rightarrow 2x = 4 \Rightarrow x = 2$

$y = 2^2 - 11(2) + 3 = -15$

$(2, -15)$

Find the unknown letter

Worked example — find a

$y = x^2 + ax + 3$ has slope $5$ when $x = 1$. Find $a$.

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

At $x = 1$, slope $= 5$: $2(1) + a = 5$

$2 + a = 5$

$a = 3$

You try

$y = x^2 + kx - 3$ has slope $12$ when $x = 3$. Find $k$.

$\tfrac{dy}{dx}=2x+k$. Put $x=3$ and the slope $12$ in.

$2(3) + k = 12$

$6 + k = 12$

$k = 6$

You try

$y = x^2 + px + 3$ has slope $10$ when $x = -2$. Find $p$.

$2(-2) + p = 10$. Mind the sign.

$-4 + p = 10$

$p = 14$

That’s Class 2.

Tangents to a curve, finding points from a slope, and unknown letters. Class 3: maximum and minimum points, increasing/decreasing, and sketching curves.