### Solved Examples

**Question 1: **What is the slope of the tangent line to the curve $y = 5x^{2} - 3x + 2$ at the point $(1, 1)$?

** Solution: **

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

Slope of the tangent to the given curve at any point $(x_{0}, y_{0})$ is given by its derivative at $x_{0}$ i.e., $\frac{dy}{dx}$$ _{x = x_{0}}$.

Differentiating with respect to $x$ on both sides of the equation,

$\frac{dy}{dx}$ $= $$\frac{d}{dx}$$(5x^{2} - 3x + 2)$

Use Power rule of derivatives $\frac{d}{dx}$ $x^{n} = n x^{n - 1}$

$\frac{dy}{dx}$ $= 10x - 3$

Plug $x = 1$ to get the slope at $(1,1)$

$\frac{dy}{dx}$$ _{x = 1}$ $= 10(1) - 3 = 7$

So, slope of the tangent to the curve $y = 5x^{2} - 3x + 2$ at $(1, 1)$ is $7$.

**Question 2: **What is the x coordinate of the point at which the curve $y = 2x^{3} - 7$ has a slope $12$?

** Solution: **

Given $y = 2x^{3} - 7$

Slope of the tangent to the given curve at any point $(x_{0}, y_{0})$ is given by its derivative at $x_{0}$ i.e., $\frac{dy}{dx}$$ _{x = x_{0}}$

Differentiating with respect to $x$ on both sides of the equation,

$\frac{dy}{dx}$$ = $$\frac{d}{dx}$$ (2x^{3} - 7)$

Use Power rule of derivatives $\frac{d}{dx}$ $x^{n} = n x^{n - 1}$

$\frac{dy}{dx}$ $= 6x$

Given the slope at a point $x = x_{0}$ is $12$,

$\frac{dy}{dx}$$ _{x = x_{0}}$ $= 6x_{0}$

$12 = 6x_{0}$

Dividing both sides by $6$,

$x_{0} = $$\frac{12}{6}$$ = 2$

Thus, the x coordinate of the point at which the curve $y = 2x^{3} - 7$ has a slope $12$ is $2$.

**Question 3: **Find the equation of the tangent line to the curve $y = 2x^{2} - 7x + 4$ at the point $(2, 4)$.

** Solution: **

Given $y = 2x^{2} - 7x + 4$

Slope of the tangent to the given curve at any point $(x_{0}, y_{0})$ is given by its derivative at $x_{0}$ i.e., $\frac{dy}{dx}$$ _{x = x_{0}}$

Differentiating with respect to $x$ on both sides of the equation,

$\frac{dy}{dx}$$ = $$\frac{d}{dx}$$ (2x^{2} - 7x + 4)$

Use Power rule of derivatives $\frac{d}{dx}$ $x^{n} = n x^{n - 1}$

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

Plug $x = 1$ to get the slope at $(2, 4)$

$\frac{dy}{dx}$$ _{x = 1}$ $= 4(2) - 7 = 1$

So, slope of the tangent to the curve $y = 2x^{2} - 7x + 4$ at $(2, 4)$ is $1$.

Use the point slope form $y - y_{0} = m(x - x_{0})$ and plug the values $x_{0} = 2, y_{0} = 4$ and $m = 1$.

$y - 4 = 1(x - 2)$

$y - 4 = x - 2$

Adding $4$ to both sides of the equation,

$y = x - 2 + 4 = x + 2$

Thus, the required tangent line equation is $y = x + 2$