Tangents and Normals

Question 1 of 7

Find the equation of the tangent line that touches
the curve

f (x) = x^{4} - 3 x^{3} + 2 x + 5

$f(x)=x^4-3x^3+2x+5$
at the point

(2, 1)

$(2,1)$

1. $y = - 2 x + 4$ $y=-2x+4$
2. $y = - 2 x + 5$ $y=-2x+5$
3. $y = x + 3$ $y=x+3$
4. $y = 2 x - 5$ $y=2x-5$

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The gradient of a tangent line to a curve

f (x)

$f(x)$ at point $(x_{1}, y_{1})$ $(x_1,y_1)$ can be found by solving for

f' (

$f'($ $x_{1}$ $x_1$

)

$)$

Point-Gradient Formula

y -

$y-$ $y_{1}$ $y_1$

=

$=$ $m$ $m$

(x -

$(x-$ $x_{1}$ $x_1$

)

$)$

First, find the gradient ( $m$ $m$ ) of the tangent line by solving for

f' (

$f'($ $2$ $2$

)

$)$ , where $2$ $2$ is the

x

$x$ -value from the given point $(2, 1)$ $(2,1)$

$f (x)$ $f(x)$	$=$ $=$	$x^{4} - 3 x^{3} + 2 x + 5$ $x^4-3x^3+2x+5$
$f' (x)$ $f'(x)$	$=$ $=$	$4 x^{3} - 9 x^{2} + 2$ $4x^3-9x^2+2$
$f' ($ $f'($ $2$ $2$ $)$ $)$	$=$ $=$	$4(\color{#00880A}{2}^3)-9(\color{#00880A}{2}^2)+2$	Substitute $x = 2$ $x=2$
	$=$ $=$	$4 (8) - 9 (4) + 2$ $4(8)-9(4)+2$
	$=$ $=$	$32 - 36 + 2$ $32-36+2$
$m$ $m$	$=$ $=$	$- 2$ $-2$

Slot the given point and the gradient into the Point-Gradient Formula

m = - 2

$m=-2$

(2, 1)

$(2,1)$

$y -$ $y-$ $y_{1}$ $y_1$	$=$ $=$	$m$ $m$ $(x -$ $(x-$ $x_{1}$ $x_1$ $)$ $)$	Point-Gradient Formula
$y -$ $y-$ $1$ $1$	$=$ $=$	$- 2$ $-2$ $(x -$ $(x-$ $2$ $2$ $)$ $)$	Substitute values
$y - 1$ $y-1$	$=$ $=$	$- 2 x + 4$ $-2x+4$
$y$ $y$	$=$ $=$	$- 2 x + 4 + 1$ $-2x+4+1$
$y$ $y$	$=$ $=$	$- 2 x + 5$ $-2x+5$

y = - 2 x + 5

$y=-2x+5$

Question 2 of 7

Find the equation of the tangent line that touches
the curve

f (x) = x^{3} + 2 x^{2} - 5 x + 2

$f(x)=x^3+2x^2-5x+2$
at

x = 0

$x=0$

1. $y = - 2 x + 5$ $y=-2x+5$
2. $y = 5 x - 4$ $y=5x-4$
3. $y = - 5 x + 2$ $y=-5x+2$
4. $y = x - 5$ $y=x-5$

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The gradient of a tangent line to a curve

f (x)

$f(x)$ at point $(x_{1}, y_{1})$ $(x_1,y_1)$ can be found by solving for

f' (

$f'($ $x_{1}$ $x_1$

)

$)$

Point-Gradient Formula

y -

$y-$ $y_{1}$ $y_1$

=

$=$ $m$ $m$

(x -

$(x-$ $x_{1}$ $x_1$

)

$)$

First, find the gradient ( $m$ $m$ ) of the tangent line by solving for

f' (

$f'($ $0$ $0$

)

$)$ (from $x = 0$ $x=0$ )

$f (x)$ $f(x)$	$=$ $=$	$x^{3} + 2 x^{2} - 5 x + 2$ $x^3+2x^2-5x+2$
$f' (x)$ $f'(x)$	$=$ $=$	$3 x^{2} + 4 x - 5$ $3x^2+4x-5$
$f' ($ $f'($ $0$ $0$ $)$ $)$	$=$ $=$	$3(\color{#00880A}{0}^2)+4(\color{#00880A}{0})-5$	Substitute $x = 0$ $x=0$
	$=$ $=$	$0 + 0 - 5$ $0+0-5$
$m$ $m$	$=$ $=$	$- 5$ $-5$

Next, find the corresponding

y

$y$ value if $x = 0$ $x=0$ is substituted to

f (x)

$f(x)$

$f (x)$ $f(x)$	$=$ $=$	$x^{3} + 2 x^{2} - 5 x + 2$ $x^3+2x^2-5x+2$
$f ($ $f($ $0$ $0$ $)$ $)$	$=$ $=$	$(\color{#00880A}{0}^3)+2(\color{#00880A}{0}^2)-5(\color{#00880A}{0})+2$	Substitute $x = 0$ $x=0$
	$=$ $=$	$0 + 0 - 0 + 2$ $0+0-0+2$
$y$ $y$	$=$ $=$	$2$ $2$

This means that the tangent line touches the curve at point $(0, 2)$ $(0,2)$

Slot this point and the gradient into the Point-Gradient Formula

m = - 5

$m=-5$

(0, 2)

$(0,2)$

$y -$ $y-$ $y_{1}$ $y_1$	$=$ $=$	$m$ $m$ $(x -$ $(x-$ $x_{1}$ $x_1$ $)$ $)$	Point-Gradient Formula
$y -$ $y-$ $2$ $2$	$=$ $=$	$- 5$ $-5$ $(x -$ $(x-$ $0$ $0$ $)$ $)$	Substitute values
$y - 2$ $y-2$	$=$ $=$	$- 5 x + 0$ $-5x+0$
$y - 2$ $y-2$ $+ 2$ $+2$	$=$ $=$	$- 5 x$ $-5x$ $+ 2$ $+2$	Add $2$ $2$ to both sides
$y$ $y$	$=$ $=$	$- 5 x + 2$ $-5x+2$

y = - 5 x + 2

$y=-5x+2$

Question 3 of 7

Find the equation of the tangent line that touches the curve

f (x) = x^{2} - 2 x + 4

$f(x)=x^2-2x+4$ at a point where the tangent line is parallel to the line

y = 2 x + 3

$y=2x+3$

1. $y = 2 x - 4$ $y=2x-4$
2. $y = 2 x$ $y=2x$
3. $y = x + 2$ $y=x+2$
4. $y = 2 x - 1$ $y=2x-1$

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The gradient of a tangent line to a curve

f (x)

$f(x)$ at point $(x_{1}, y_{1})$ $(x_1,y_1)$ can be found by solving for

f' (

$f'($ $x_{1}$ $x_1$

)

$)$

Point-Gradient Formula

y -

$y-$ $y_{1}$ $y_1$

=

$=$ $m$ $m$

(x -

$(x-$ $x_{1}$ $x_1$

)

$)$

First, find the gradient of the given parallel line

$y$ $y$	$=$ $=$	$m x + b$ $mx+b$	Gradient-Intercept Form
$y$ $y$	$=$ $=$	$2 x + 3$ $2x+3$

m_{parallel}

$m_(\text(parallel))$

=

$=$

2

$2$

Also, note that since parallel lines have equal gradients, this means that $m_{tangent} = 2$ $m_(\text(tangent))=2$

Next, find the point where the tangent line touches the curve.

Do this by equating

f' (x)

$f'(x)$ to

m_{parallel}

$m_(\text(parallel))$ , then solving for

x

$x$ and

y

$y$

$f (x)$ $f(x)$	$=$ $=$	$x^{2} - 2 x + 4$ $x^2-2x+4$
$f' (x)$ $f'(x)$	$=$ $=$	$2 x - 2 = 2$ $2x-2=2$	Equate to $m_{parallel} = 2$ $m_(\text(parallel))=2$

$2 x$ $2x$	$=$ $=$	$2 + 2$ $2+2$
$2 x$ $2x$	$=$ $=$	$4$ $4$
$2 x$ $2x$ $\div 2$ $-:2$	$=$ $=$	$4$ $4$ $\div 2$ $-:2$	Divide both sides by $2$ $2$
$x$ $x$	$=$ $=$	$2$ $2$

Substitute this

x

$x$ value to the main function to solve for the corresponding

y

$y$ value

$f (x)$ $f(x)$	$=$ $=$	$x^{2} - 2 x + 4$ $x^2-2x+4$
$f ($ $f($ $2$ $2$ $)$ $)$	$=$ $=$	$\color{#00880A}{2}^2-2(\color{#00880A}{2})+4$	Substitute $x = 2$ $x=2$
	$=$ $=$	$4 - 4 + 4$ $4-4+4$
$y$ $y$	$=$ $=$	$4$ $4$

This means that the tangent line touches the curve at $(2, 4)$ $(2,4)$

Slot the point and the gradient into the Point-Gradient Formula

m = 2

$m=2$

(2, 4)

$(2,4)$

$y -$ $y-$ $y_{1}$ $y_1$	$=$ $=$	$m$ $m$ $(x -$ $(x-$ $x_{1}$ $x_1$ $)$ $)$	Point-Gradient Formula
$y -$ $y-$ $4$ $4$	$=$ $=$	$2$ $2$ $(x -$ $(x-$ $2$ $2$ $)$ $)$	Substitute values
$y - 4$ $y-4$	$=$ $=$	$2 x - 4$ $2x-4$
$y$ $y$	$=$ $=$	$2 x - 4 + 4$ $2x-4+4$
$y$ $y$	$=$ $=$	$2 x$ $2x$

y = 2 x

$y=2x$

Question 4 of 7

Find the equation of the tangent line that touches the curve

f (x) = (x - 2) (x^{2} + 2 x + 6)

$f(x)=(x-2)(x^2+2x+6)$ at the

x

$x$ -axis

1. $y = 14 (x - 2)$ $y=14(x-2)$
2. $y = 14 x$ $y=14x$
3. $y = 14 (x - 1)$ $y=14(x-1)$
4. $y = 14 (x + 3)$ $y=14(x+3)$

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The gradient of a tangent line to a curve

f (x)

$f(x)$ at point $(x_{1}, y_{1})$ $(x_1,y_1)$ can be found by solving for

f' (

$f'($ $x_{1}$ $x_1$

)

$)$

Point-Gradient Formula

y -

$y-$ $y_{1}$ $y_1$

=

$=$ $m$ $m$

(x -

$(x-$ $x_{1}$ $x_1$

)

$)$

First, find the exact point where the tangent line touches the curve

Since the tangent intersects the

x

$x$ -axis at this point, we know that $y = 0$ $y=0$

$f (x)$ $f(x)$	$=$ $=$	$(x - 2) (x^{2} + 2 x + 6)$ $(x-2)(x^2+2x+6)$
$0$ $0$	$=$ $=$	$(x - 2) (x^{2} + 2 x + 6)$ $(x-2)(x^2+2x+6)$	Substitute $y = 0$ $y=0$

$\frac{0}{\color{#CC0000}{x^2+2x+6}}$	$=$ $=$	$\frac{(x-2)(x^2+2x+6)}{\color{#CC0000}{x^2+2x+6}}$	Divide both sides by $x^{2} + 2 x + 6$ $x^2+2x+6$

$0$ $0$	$=$ $=$	$x - 2$ $x-2$
$2$ $2$	$=$ $=$	$x$ $x$
$x$ $x$	$=$ $=$	$2$ $2$

This means that the tangent line touches the curve at point $(2, 0)$ $(2,0)$

Next, find the gradient ( $m$ $m$ ) of the tangent line by solving for

f' (

$f'($ $2$ $2$

)

$)$ (from $x = 2$ $x=2$ )

$f (x)$ $f(x)$	$=$ $=$	$(x - 2) (x^{2} + 2 x + 6)$ $(x-2)(x^2+2x+6)$
	$=$ $=$	$x^{3} - 2 x^{2} + 2 x^{2} - 4 x + 6 x - 12$ $x^3-2x^2+2x^2-4x+6x-12$	Distribute $(x - 2)$ $(x-2)$
	$=$ $=$	$x^{3} + 2 x - 12$ $x^3+2x-12$
$f' (x)$ $f'(x)$	$=$ $=$	$3 x^{2} + 2$ $3x^2+2$
$f' ($ $f'($ $2$ $2$ $)$ $)$	$=$ $=$	$3(\color{#00880A}{2}^2)+2$	Substitute $x = 2$ $x=2$
	$=$ $=$	$12 + 2$ $12+2$
$m$ $m$	$=$ $=$	$14$ $14$

Slot the point and the gradient into the Point-Gradient Formula

m = 14

$m=14$

(2, 0)

$(2,0)$

$y -$ $y-$ $y_{1}$ $y_1$	$=$ $=$	$m$ $m$ $(x -$ $(x-$ $x_{1}$ $x_1$ $)$ $)$	Point-Gradient Formula
$y -$ $y-$ $0$ $0$	$=$ $=$	$14$ $14$ $(x -$ $(x-$ $2$ $2$ $)$ $)$	Substitute values
$y$ $y$	$=$ $=$	$14 (x - 2)$ $14(x-2)$

y = 14 (x - 2)

$y=14(x-2)$

Question 5 of 7

Find the equation of the normal line that intersects the curve

f (x) = {(x + 2)}^{2}

$f(x)=(x+2)^2$ at point

(- 3, 1)

$(-3,1)$

1. $2 x - 2 y - 5 = 0$ $2x-2y-5=0$
2. $2 x - y + 5 = 0$ $2x-y+5=0$
3. $x + y - 5 = 0$ $x+y-5=0$
4. $x - 2 y + 5 = 0$ $x-2y+5=0$

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The gradient of a tangent line to a curve

f (x)

$f(x)$ at point $(x_{1}, y_{1})$ $(x_1,y_1)$ can be found by solving for

f' (

$f'($ $x_{1}$ $x_1$

)

$)$

Point-Gradient Formula

y -

$y-$ $y_{1}$ $y_1$

=

$=$ $m$ $m$

(x -

$(x-$ $x_{1}$ $x_1$

)

$)$

A normal line to a curve at a point is perpendicular to the tangent line at the same point.

First, find the gradient ( $m$ $m$ ) of the tangent line by solving for

f' (

$f'($ $- 3$ $-3$

)

$)$ , where $- 3$ $-3$ is the

x

$x$ -value from the given point $(- 3, 1)$ $(-3,1)$

$f (x)$ $f(x)$	$=$ $=$	${(x + 2)}^{2}$ $(x+2)^2$
$f' (x)$ $f'(x)$	$=$ $=$	$2 (x + 2)$ $2(x+2)$
$f' ($ $f'($ $- 3$ $-3$ $)$ $)$	$=$ $=$	$2(\color{#00880A}{-3}+2)$	Substitute $x = - 3$ $x=-3$
	$=$ $=$	$2 (- 1)$ $2(-1)$
$m_{tangent}$ $m_(\text(tangent))$	$=$ $=$	$- 2$ $-2$

Remember that the gradients of perpendicular lines are negative reciprocals of each other

Find the gradient of the normal line by getting the negative reciprocal of the gradient of the tangent line

$m_{tangent}$ $m_(\text(tangent))$	$=$ $=$	$- 2$ $-2$

	$=$ $=$	$- \frac{1}{2}$ $-1/2$	Reciprocate

	$=$ $=$	$- \frac{1}{2} \times - 1$ $-1/2xx-1$	Multiply by $- 1$ $-1$

$m_{normal}$ $m_(\text(normal))$	$=$ $=$	$\frac{1}{2}$ $1/2$

Slot the given point and the gradient into the Point-Gradient Formula

m = \frac{1}{2}

$m=1/2$

(- 3, 1)

$(-3,1)$

$y -$ $y-$ $y_{1}$ $y_1$	$=$ $=$	$m$ $m$ $(x -$ $(x-$ $x_{1}$ $x_1$ $)$ $)$	Point-Gradient Formula
$y -$ $y-$ $1$ $1$	$=$ $=$	$\frac{1}{2}$ $1/2$ $(x - ($ $(x-($ $- 3$ $-3$ $))$ $))$	Substitute values

$(y - 1)$ $(y-1)$ $\times 2$ $xx2$	$=$ $=$	$\frac{1}{2} (x + 3)$ $1/2(x+3)$ $\times 2$ $xx2$	Multiply $2$ $2$ to both sides

$2 y - 2$ $2y-2$	$=$ $=$	$x + 3$ $x+3$
$0$ $0$	$=$ $=$	$x + 3 - 2 y + 2$ $x+3-2y+2$
$0$ $0$	$=$ $=$	$x - 2 y + 5$ $x-2y+5$
$x - 2 y + 5$ $x-2y+5$	$=$ $=$	$0$ $0$

x - 2 y + 5 = 0

$x-2y+5=0$

Question 6 of 7

Find the equation of the normal line that intersects
the curve

f (x) = x^{3} + 2 x^{2} - 5 x

$f(x)=x^3+2x^2-5x$
at the point

(- 3, 6)

$(-3,6)$

1. $x - 10 y - 60 = 0$ $x-10y-60=0$
2. $x + 10 y - 57 = 0$ $x+10y-57=0$
3. $x - 5 y - 57 = 0$ $x-5y-57=0$
4. $x + 5 y - 60 = 0$ $x+5y-60=0$

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The gradient of a tangent line to a curve

f (x)

$f(x)$ at point $(x_{1}, y_{1})$ $(x_1,y_1)$ can be found by solving for

f' (

$f'($ $x_{1}$ $x_1$

)

$)$

Point-Gradient Formula

y -

$y-$ $y_{1}$ $y_1$

=

$=$ $m$ $m$

(x -

$(x-$ $x_{1}$ $x_1$

)

$)$

A normal line to a curve at a point is perpendicular to the tangent line at the same point.

First, find the gradient ( $m$ $m$ ) of the tangent line by solving for

f' (

$f'($ $- 3$ $-3$

)

$)$ , where $- 3$ $-3$ is the

x

$x$ -value from the given point $(- 3, 6)$ $(-3,6)$

$f (x)$ $f(x)$	$=$ $=$	$x^{3} + 2 x^{2} - 5 x$ $x^3+2x^2-5x$
$f' (x)$ $f'(x)$	$=$ $=$	$3 x^{2} + 4 x - 5$ $3x^2+4x-5$
$f' ($ $f'($ $- 3$ $-3$ $)$ $)$	$=$ $=$	$3(\color{#00880A}{-3}^2)+4(\color{#00880A}{-3})-5$	Substitute $x = - 3$ $x=-3$
	$=$ $=$	$27 - 12 - 5$ $27-12-5$
$m_{tangent}$ $m_(\text(tangent))$	$=$ $=$	$10$ $10$

Remember that the gradients of perpendicular lines are negative reciprocals of each other

Find the gradient of the normal line by getting the negative reciprocal of the gradient of the tangent line

$m_{tangent}$ $m_(\text(tangent))$	$=$ $=$	$10$ $10$

	$=$ $=$	$\frac{1}{10}$ $1/10$	Reciprocate

	$=$ $=$	$\frac{1}{10} \times - 1$ $1/10xx-1$	Multiply by $- 1$ $-1$

$m_{normal}$ $m_(\text(normal))$	$=$ $=$	$- \frac{1}{10}$ $-1/10$

Slot the given point and the gradient into the Point-Gradient Formula

m = - \frac{1}{10}

$m=-1/10$

(- 3, 6)

$(-3,6)$

$y -$ $y-$ $y_{1}$ $y_1$	$=$ $=$	$m$ $m$ $(x -$ $(x-$ $x_{1}$ $x_1$ $)$ $)$	Point-Gradient Formula
$y -$ $y-$ $6$ $6$	$=$ $=$	$- \frac{1}{10}$ $-1/10$ $(x - ($ $(x-($ $- 3$ $-3$ $))$ $))$	Substitute values

$(y - 6)$ $(y-6)$ $\times 10$ $xx10$	$=$ $=$	$- \frac{1}{10} (x + 3)$ $-1/10(x+3)$ $\times 10$ $xx10$	Multiply $10$ $10$ to both sides

$10 y - 60$ $10y-60$	$=$ $=$	$- x - 3$ $-x-3$
$10 y - 60 + x + 3$ $10y-60+x+3$	$=$ $=$	$0$ $0$
$x + 10 y - 57$ $x+10y-57$	$=$ $=$	$0$ $0$

x + 10 y - 57 = 0

$x+10y-57=0$

Question 7 of 7

A tangent and normal line both touch the curve

f (x) = x^{3} - x^{2} - 6

$f(x)=x^3-x^2-6$ at the point

(2, - 2)

$(2,-2)$ .
However, these lines intersect the

x

$x$ axis at different points.
Find the distance between these two points.

1. $10 \frac{1}{4}$ $10 1/4$
2. $16 \frac{2}{3}$ $16 2/3$
3. $10 \frac{2}{3}$ $10 2/3$
4. $16 \frac{1}{4}$ $16 1/4$

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The gradient of a tangent line to a curve

f (x)

$f(x)$ at point $(x_{1}, y_{1})$ $(x_1,y_1)$ can be found by solving for

f' (

$f'($ $x_{1}$ $x_1$

)

$)$

Point-Gradient Formula

y -

$y-$ $y_{1}$ $y_1$

=

$=$ $m$ $m$

(x -

$(x-$ $x_{1}$ $x_1$

)

$)$

A normal line to a curve at a point is perpendicular to the tangent line at the same point.

Form the equation of the tangent line

Start with finding the gradient ( $m_{tangent}$ $m_(\text(tangent))$ ) of the tangent line by solving for

f' (

$f'($ $2$ $2$

)

$)$ , where $2$ $2$ is the

x

$x$ -value from the plotted point $(2, - 2)$ $(2,-2)$

$f (x)$ $f(x)$	$=$ $=$	$x^{3} - x^{2} - 6$ $x^3-x^2-6$
$f' (x)$ $f'(x)$	$=$ $=$	$3 x^{2} - 2 x$ $3x^2-2x$
$f' ($ $f'($ $2$ $2$ $)$ $)$	$=$ $=$	$3(\color{#00880A}{2}^2)-2(\color{#00880A}{2})$	Substitute $x = 2$ $x=2$
	$=$ $=$	$12 - 4$ $12-4$
$m_{tangent}$ $m_(\text(tangent))$	$=$ $=$	$8$ $8$

Slot the given point and the gradient into the Point-Gradient Formula

m = 8

$m=8$

(2, - 2)

$(2,-2)$

$y -$ $y-$ $y_{1}$ $y_1$	$=$ $=$	$m$ $m$ $(x -$ $(x-$ $x_{1}$ $x_1$ $)$ $)$	Point-Gradient Formula
$y - ($ $y-($ $- 2$ $-2$ $)$ $)$	$=$ $=$	$8$ $8$ $(x -$ $(x-$ $2$ $2$ $)$ $)$	Substitute values
$y + 2$ $y+2$	$=$ $=$	$8 x - 16$ $8x-16$
$y$ $y$	$=$ $=$	$8 x - 16 - 2$ $8x-16-2$
$y$ $y$	$=$ $=$	$8 x - 18$ $8x-18$

This is the equation of the tangent line

Identify the point where the tangent intersects the

x

$x$ axis by substituting

y = 0

$y=0$ into the tangent equation

$y$ $y$	$=$ $=$	$8 x - 18$ $8x-18$
$0$ $0$	$=$ $=$	$8 x - 18$ $8x-18$	Substitute $y = 0$ $y=0$
$18$ $18$	$=$ $=$	$8 x$ $8x$
$18$ $18$ $\div 8$ $-:8$	$=$ $=$	$8 x$ $8x$ $\div 8$ $-:8$	Divide both sides by $8$ $8$

$\frac{9}{4}$ $9/4$	$=$ $=$	$x$ $x$

$x_{tangent}$ $x_(\text(tangent))$	$=$ $=$	$\frac{9}{4}$ $9/4$ or $2 \frac{1}{4}$ $2 1/4$

The tangent line intersects the

x

$x$ axis at $x = 2 \frac{1}{4}$ $x=2 1/4$

Now, form the equation of the normal line

Start with finding the gradient ( $m_{normal}$ $m_(\text(normal))$ ) of the normal line by getting the negative reciprocal of the gradient of the tangent line

$m_{tangent}$ $m_(\text(tangent))$	$=$ $=$	$8$ $8$

	$=$ $=$	$\frac{1}{8}$ $1/8$	Reciprocate

	$=$ $=$	$\frac{1}{8} \times - 1$ $1/8xx-1$	Multiply by $- 1$ $-1$

$m_{normal}$ $m_(\text(normal))$	$=$ $=$	$- \frac{1}{8}$ $-1/8$

Slot the given point and the gradient into the Point-Gradient Formula

m = - \frac{1}{8}

$m=-1/8$

(2, - 2)

$(2,-2)$

$y -$ $y-$ $y_{1}$ $y_1$	$=$ $=$	$m$ $m$ $(x -$ $(x-$ $x_{1}$ $x_1$ $)$ $)$	Point-Gradient Formula
$y - ($ $y-($ $- 2$ $-2$ $)$ $)$	$=$ $=$	$- \frac{1}{8}$ $-1/8$ $(x -$ $(x-$ $2$ $2$ $)$ $)$	Substitute values

$(y + 2)$ $(y+2)$ $\times 8$ $xx8$	$=$ $=$	$- \frac{1}{8} (x - 2)$ $-1/8(x-2)$ $\times 8$ $xx8$	Multiply $8$ $8$ to both sides

$8 y + 16$ $8y+16$	$=$ $=$	$- x + 2$ $-x+2$
$8 y + 16 + x - 2$ $8y+16+x-2$	$=$ $=$	$0$ $0$
$x + 8 y + 14$ $x+8y+14$	$=$ $=$	$0$ $0$

This is the equation of the normal line

Identify the point where the normal intersects the

x

$x$ axis by substituting

y = 0

$y=0$ into the normal equation

$x + 8 y + 14$ $x+8y+14$	$=$ $=$	$0$ $0$
$x + 8 (0) + 14$ $x+8(0)+14$	$=$ $=$	$0$ $0$	Substitute $y = 0$ $y=0$
$x + 14$ $x+14$	$=$ $=$	$0$ $0$
$x + 14$ $x+14$ $- 14$ $-14$	$=$ $=$	$0$ $0$ $- 14$ $-14$	Subtract $14$ $14$ from both sides
$x_{normal}$ $x_(\text(normal))$	$=$ $=$	$- 14$ $-14$

The normal line intersects the

x

$x$ axis at $x = - 14$ $x=-14$

Finally, get the distance between the two

x

$x$ values by subtraction

$x_{tangent} - x_{normal}$ $x_(\text(tangent))-x_(\text(normal))$	$=$ $=$	$2 \frac{1}{4} - (- 14)$ $2 1/4-(-14)$

	$=$ $=$	$2 \frac{1}{4} + 14$ $2 1/4+14$

	$=$ $=$	$16 \frac{1}{4}$ $16 1/4$

16 \frac{1}{4}

$16 1/4$

Tangents and Normals

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Quiz summary

Information

1. Question

Hint

Point-Gradient Formula

2. Question

Hint

Point-Gradient Formula

3. Question

Hint

Point-Gradient Formula

4. Question

Hint

Point-Gradient Formula

5. Question

Point-Gradient Formula

6. Question

Hint

Point-Gradient Formula

7. Question

Hint

Point-Gradient Formula

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