# How do you find the slope of the tangent line to the graph of the function h(t)=t^2+3 at (-2,7)?

Jun 30, 2017

The equation of the tangent is -

$y = - 4 x - 1$

#### Explanation:

Given -

$h \left(t\right) = {t}^{2} + 3$

Let us have it as -

$y = {t}^{2} + 3$

Its slope at any point is given by its first derivative.

$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 t$

Slope of the curve exactly at $x = - 2$ is-

$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 \left(- 2\right) = - 4$

The tangent is passing through the point $\left(- 2 , 7\right)$

The slope of the tangent at Point $x = - 2$ is the same as slope of the curve at that point $\left(- 2 , 7\right)$

$m = - 4$
$x = - 2$
$y = 7$
$m x + c = y$
$\left(- 4\right) \left(- 2\right) + c = 7$
$8 + c = 7$
$c = 7 - 8 = - 1$

The equation of the tangent is -

$y = - 4 x - 1$