Question

Find an equation of the tangent line to the graph of y = x2 + 3x at the point (-1, -2)

show clear work plz

Answer 1

We have to find an equation of the tangent line to the graph of

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

at the point (-1,-2).

Now, the equation of the tangent line at the point (1,-2) is

$y-(-2)=\left | \frac{dy}{dx} \right |_{-1,-2}(x-(-1))$

Now, from the equation of the graph, we find the derivative.

Now,

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

$or, \frac{dy}{dx}=2x+3$

So, at the point (-1,-2), the value of the derivative becomes

$\frac{dy}{dx}_{(-1,-2)}=2(-1)+3=1$

So, the slope of the straight line is 1.

This means, the equation of the tangent line is

$y-(-2)=1[x-(-1)]$

Thus, the equation is x-y=1.

The answer is

The equation of the tangent line to the graph of

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

at the point (-1,-2) is given by the equation

$x-y=1$.