Question

find the equation of tangent to the curve y=x^2-4x-5 at point x= -2​

Answer 1

$\large\underline{\sf{Solution-}}$

Given curve is

$\rm :\longmapsto\:y = {x}^{2} - 4x - 5$

Now, when x = -2, then

$\rm :\longmapsto\:y = {( - 2)}^{2} - 4( - 2) - 5$

$\rm :\longmapsto\:y = 4 + 8 - 5$

$\rm :\longmapsto\:y = 12 - 5$

$\rm :\longmapsto\:y = 7$

So, required point is (- 2, 7) at which we have to find equation of tangent.

So, let first find slope of tangent.

We have

$\rm :\longmapsto\:y = {x}^{2} - 4x - 5$

Differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx} y =\dfrac{d}{dx}( {x}^{2} - 4x - 5)$

$\rm :\longmapsto\:\dfrac{dy}{dx} =\dfrac{d}{dx}{x}^{2} -\dfrac{d}{dx} 4x -\dfrac{d}{dx} 5$

$\rm :\longmapsto\:\dfrac{dy}{dx} =\dfrac{d}{dx}{x}^{2} -4\dfrac{d}{dx}x -\dfrac{d}{dx} 5$

We know,

$\underbrace{ \boxed{ \bf \:\dfrac{d}{dx} {x}^{n} = {nx}^{n - 1}}}$

$\underbrace{ \boxed{ \bf \:\dfrac{d}{dx} {x} = 1}}$

$\underbrace{ \boxed{ \bf \:\dfrac{d}{dx} {k} = 0}}$

Now, using all these results, we get

$\rm :\longmapsto\:\dfrac{dy}{dx} = 2x - 4$

Now, we know that

If y = f(x) be any curve then slope of tangent at point P is represented by m and is given by

$\underbrace{ \boxed{ \bf \: Slope \: of \: tangent \: = \bigg(\dfrac{dy}{dx}\bigg) _P}}$

So,

Slope of tangent is

$\rm :\longmapsto\: \: Slope \: of \: tangent ,\: m = \bigg(\dfrac{dy}{dx}\bigg) _{( - 2,7)}$

$\rm \: = \: 2( - 2) - 4$

$\rm \: = \: - 4 - 4$

$\rm \: = \: - 8$

So,

Slope of tangent, at point (- 2, 7), m = - 8.

Now, we know that

Equation of line which passes through the point (a, b) having slope m, using slope point form is given by

$\underbrace{ \boxed{ \bf \: y - b = m(x - a)}}$

Hence,

Equation of tangent at point (- 2, 7) having slope m = - 8 is given by

$\rm :\longmapsto\:y - 7 = - 8(x + 2)$

$\rm :\longmapsto\:y - 7 = - 8x - 16$

$\rm :\longmapsto\:8x + y = - 16 + 7$

$\rm :\longmapsto\:8x + y = - 9$

$\rm :\longmapsto\:8x + y + 9 = 0$

Additional Information :-

Let y = f(x) be any curve, then line which touches the curve y = f(x) exactly at one point say P is called tangent and at that very point P, if we draw a perpendicular on tangent, that line is called normal to the curve at P.

2. If tangent line is parallel to x - axis, its slope is 0.

3. If tangent line is parallel to y - axis, its slope is not defined

4. Two lines having slope M and m are parallel, iff M = m

5. If two lines having slope M and m are perpendicular, iff Mm = - 1.

Answer 2

Answer:

$\large\underline{\sf{Solution-}}Solution−</p><p></p><p>Given curve is</p><p></p><p>\rm :\longmapsto\:y = {x}^{2} - 4x - 5:⟼y=x2−4x−5</p><p></p><p>Now, when x = -2, then</p><p></p><p>\rm :\longmapsto\:y = {( - 2)}^{2} - 4( - 2) - 5:⟼y=(−2)2−4(−2)−5</p><p></p><p>\rm :\longmapsto\:y = 4 + 8 - 5:⟼y=4+8−5</p><p></p><p>\rm :\longmapsto\:y = 12 - 5:⟼y=12−5</p><p></p><p>\rm :\longmapsto\:y = 7:⟼y=7</p><p></p><p>So, required point is (- 2, 7) at which we have to find equation of tangent.</p><p></p><p>So, let first find slope of tangent.</p><p></p><p>We have</p><p></p><p>\rm :\longmapsto\:y = {x}^{2} - 4x - 5:⟼y=x2−4x−5</p><p></p><p>Differentiating both sides w. r. t. x, we get</p><p></p><p>\rm :\longmapsto\:\dfrac{d}{dx} y =\dfrac{d}{dx}( {x}^{2} - 4x - 5):⟼dxdy=dxd(x2−4x−5)</p><p></p><p>\rm :\longmapsto\:\dfrac{dy}{dx} =\dfrac{d}{dx}{x}^{2} -\dfrac{d}{dx} 4x -\dfrac{d}{dx} 5:⟼dxdy=dxdx2−dxd4x−dxd5</p><p></p><p>\rm :\longmapsto\:\dfrac{dy}{dx} =\dfrac{d}{dx}{x}^{2} -4\dfrac{d}{dx}x -\dfrac{d}{dx} 5:⟼dxdy=dxdx2−4dxdx−dxd5</p><p></p><p>We know,</p><p></p><p>\underbrace{ \boxed{ \bf \:\dfrac{d}{dx} {x}^{n} = {nx}^{n - 1}}}dxdxn=nxn−1</p><p></p><p>\underbrace{ \boxed{ \bf \:\dfrac{d}{dx} {x} = 1}}dxdx=1</p><p></p><p>\underbrace{ \boxed{ \bf \:\dfrac{d}{dx} {k} = 0}}dxdk=0</p><p></p><p>Now, using all these results, we get</p><p></p><p>\rm :\longmapsto\:\dfrac{dy}{dx} = 2x - 4:⟼dxdy=2x−4</p><p></p><p>Now, we know that</p><p></p><p>If y = f(x) be any curve then slope of tangent at point P is represented by m and is given by</p><p></p><p>\underbrace{ \boxed{ \bf \: Slope \: of \: tangent \: = \bigg(\dfrac{dy}{dx}\bigg) _P}}Slopeoftangent=(dxdy)P</p><p></p><p>So,</p><p></p><p>Slope of tangent is</p><p></p><p>\rm :\longmapsto\: \: Slope \: of \: tangent ,\: m = \bigg(\dfrac{dy}{dx}\bigg) _{( - 2,7)}:⟼Slopeoftangent,m=(dxdy)(−2,7)</p><p></p><p>\rm \: = \: 2( - 2) - 4=2(−2)−4</p><p></p><p>\rm \: = \: - 4 - 4=−4−4</p><p></p><p>\rm \: = \: - 8=−8</p><p></p><p>So,</p><p></p><p>Slope of tangent, at point (- 2, 7), m = - 8.</p><p></p><p>Now, we know that</p><p></p><p>Equation of line which passes through the point (a, b) having slope m, using slope point form is given by</p><p></p><p>\underbrace{ \boxed{ \bf \: y - b = m(x - a)}}y−b=m(x−a)</p><p></p><p>Hence,</p><p></p><p>Equation of tangent at point (- 2, 7) having slope m = - 8 is given by</p><p></p><p>\rm :\longmapsto\:y - 7 = - 8(x + 2):⟼y−7=−8(x+2)</p><p></p><p>\rm :\longmapsto\:y - 7 = - 8x - 16:⟼y−7=−8x−16</p><p></p><p>\rm :\longmapsto\:8x + y = - 16 + 7:⟼8x+y=−16+7</p><p></p><p>\rm :\longmapsto\:8x + y = - 9:⟼8x+y=−9</p><p></p><p>\rm :\longmapsto\:8x + y + 9 = 0:⟼8x+y+9=0</p><p></p><p>Additional Information :-</p><p></p><p>Let y = f(x) be any curve, then line which touches the curve y = f(x) exactly at one point say P is called tangent and at that very point P, if we draw a perpendicular on tangent, that line is called normal to the curve at P.</p><p></p><p>2. If tangent line is parallel to x - axis, its slope is 0.</p><p></p><p>3. If tangent line is parallel to y - axis, its slope is not defined</p><p></p><p>4. Two lines having slope M and m are parallel, iff M = m</p><p></p><p>5. If two lines having slope M and m are perpendicular, iff Mm = - 1.</p><p></p><p>$