Question

Find the slope of tangent of y = x^2+3x-3 at the point (1,0)​

Answer 1

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

Given curve is

$\rm :\longmapsto\:y = {x}^{2} + 3x - 3$

On differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx}y =\dfrac{d}{dx}\bigg[ {x}^{2} + 3x - 3\bigg] \:$

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{d}{dx} {x}^{2} + 3\dfrac{d}{dx}x - \dfrac{d}{dx}3$

We know,

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

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

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

So, using this, we get

$\rm :\longmapsto\:\dfrac{dy}{dx} = 2 {x}^{2 - 1} + 3 - 0$

$\rm \implies\:\boxed{\tt{ \dfrac{dy}{dx} = 2x + 3 \: }}$

Now, we know that slope of tangent to the curve y = f(x) at the point P is given by

$\boxed{\tt{ Slope \: of \: tangent, \: m = \bigg[\dfrac{dy}{dx} \bigg]_P}}$

So, slope of tangent at the point (1, 0) is

$\rm :\longmapsto\: Slope \: of \: tangent, \: m = \bigg[\dfrac{dy}{dx} \bigg]_P$

$\rm \:  =  \: 2(1) + 3$

$\rm \:  =  \: 5$

Now, we know that

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

$\rm :\longmapsto\:\boxed{\tt{ y - b) \: = \: m(x - a) \: }}$

So, equation of tangent to the curve at the point (1, 0) having slope m = 5 is given by

$\rm :\longmapsto\:y - 0 = 5(x - 1)$

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

$\bf\implies \:5x - y - 5 = 0$

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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 that very point P, if we draw a perpendicular on tangent, that line is called normal to the curve at P.

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

3. If tangent 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.