Equation of a tangent line

Assume that the tangent line equation is in the form $y=mx+q$

We can use this formula: $m=2ax_0+b$ to find $m$ .

$m=-2*2*1+0=-4$

Then we use the equation of a line $y=mx+q$ to find $q$

$q=y-mx=13+4=17$

So, having both $m$ and $q$ we can use them to write the equation of the tangent:

$\boxed{y=-4x+17}$

The slope of the curve $y=f(x)$ at any point is its first derivative $(\dfrac{dy}{dx})$ .

We can write $f(x)=15-2x^2$ as $y=15-2x^2$

Taking the derivative with respect to $x$ , we get

$\dfrac{dy}{dx}=-4x$

The slope at $x=1$ is

$\dfrac{dy}{dx}=-4(1)=-4$

It follows that,

$y=15-2(1^2)=15-2=13$

From point-slope form of an equation of a line, we have

$y-y_1=m(x-x_1)$

$y-13=-4(x-1)$

$y-13=-4x+4$

$y=-4x+4+13$

$y=-4x+17$ $\boxed{\text{answer}}$