First find the slope of the tangent to the line by taking the derivative.

Using the Exponential Rule we get the following,

\(\displaystyle y=3x^3\) 

 \(\displaystyle \frac{dy}{dx}=9x^2\).

Then plug 1 into the equation as 1 is the point to find the slope at. 

\(\displaystyle \frac{dy}{dx}=9(1)^2=9(1)=9\).

One way of finding the slope at a given point is by finding the derivative. In this case, we can take the derivative of y with respect to x, and plug in the desired value for x.

\(\displaystyle y=x^3+2x^2+x\)

Using the exponential rule we get the following derivative,

\(\displaystyle \frac{\mathrm{d}y }{\mathrm{d} x}=3x^2 +4x+1\).

Plugging in x=2 from the point 2,3 gives us the final slope,

\(\displaystyle 3(2)^2+4(2)+1\)

\(\displaystyle =3(4)+8+1\)

\(\displaystyle =12+9=21\)

Thus our slope at the specific point is \(\displaystyle 21\).

Note that in this case, using the y coordinate was not necessary.

To find the slope of the tangent line of the function at the given value, evaluate the first derivative for the given.

The first derivative is

\(\displaystyle f^{'}(x)=anx^{n-1}\)

and for this function

\(\displaystyle f^{'}(x)=3(2)x^{2-1}=6x\)

and

\(\displaystyle f'(3)=6(3)=18\)

So the slope is

\(\displaystyle 18\)

To find the slope of the tangent line of the function at the given value, evaluate the first derivative for the given value.

The first derivative is

\(\displaystyle f^{'}(x)=anx^{n-1}\)

and for this function

\(\displaystyle f^{'}(x)=2x^{2-1}+4x^{1-1}=2x+4\)

and plugging in the specific x value we get,

\(\displaystyle f'(1)=2(1)+4=6\)

So the slope is

\(\displaystyle 6\).

Calculate the derivative of \(\displaystyle y=2x^2+10x\) by using the derivative rules.  The derivative function determines the slope at any point of the original function.

The derivative is: \(\displaystyle y'=4x+10\)

With the given point \(\displaystyle (1,12)\), \(\displaystyle x=1\).   Substitute this value to the derivative function to determine the slope at that point.

\(\displaystyle y'=4(1)+10=14\)

The slope of the tangent line that intersects point \(\displaystyle (1,12)\) is \(\displaystyle 14\).

We begin by recalling that one way of defining the derivative of a function is the slope of the tangent line of the function at a given point.  Therefore, finding the derivative of our equation will allow us to find the slope of the tangent line.  Since the two things needed to find the equation of a line are the slope and a point, we would be halfway done.

We calculate the derivative using the power rule.

\(\displaystyle f'(x)=3(-4x^2)+2(7x)-9=-12x^2+14x-9\)

However, we don't want the slope of the tangent line at just any point but rather specifically at the point \(\displaystyle (1,6)\).  To obtain this, we simply substitute our x-value 1 into the derivative.

\(\displaystyle f'(1)=-12(1^2)+14(1)-9=-12+14-9=-7\)

Therefore, the slope of our tangent line is \(\displaystyle -7\).

We now need a point on our tangent line.  Our choices are quite limited, as the only point on the tangent line that we know is the point where it intersects our original graph, namely the point \(\displaystyle (1,6)\).

Therefore, we can plug these coordinates along with our slope into the general point-slope form to find the equation.

\(\displaystyle y-y_1=m(x-x_1)\)

\(\displaystyle y-6=-7(x-1)\)

Solving for \(\displaystyle y\) will give us our slope-intercept form.

\(\displaystyle y-6=-7x+7\)

\(\displaystyle y=-7x+13\)

The equation of the tangent line at \(\displaystyle \small x=2\) depends on the derivative at that point and the function value.

The derivative at that point of \(\displaystyle \small f(x)\) is 

\(\displaystyle \small f'(x)=3(x-2)^2\) using the Power Rule

\(\displaystyle f(x)=x^n \rightarrow f'(x)=nx^{n-1}\)

which means

\(\displaystyle \small f'(2)=0\)

The derivative is zero, so the tangent line will be horizontal.

It intersects it at \(\displaystyle \small (2,1)\) since \(\displaystyle \small f(x)=(x-2)^3+1 \rightarrow f(2)=(2-2)^3+1 \rightarrow f(2)=1\), so that line is \(\displaystyle \small y=1\).

Rewrite \(\displaystyle 4x-2y=12\) in slope-intercept form, \(\displaystyle y=mx+b\), to determine the slope.

\(\displaystyle 4x=12+2y\)

\(\displaystyle 4x-12=2y\)

\(\displaystyle y=2x-6\)

The slope of the given function is 2.

Substitute the slope and the given point, \(\displaystyle (1,20)\), in the slope-intercept form to determine the y-intercept.

\(\displaystyle 20=(2)(1)+b\)

\(\displaystyle b=18\)

Substitute this and the slope back to the slope-intercept equation.

The equation of the tangent line is:  \(\displaystyle y=2x+18\)

We begin by finding the equation of the derivative using the limit definition:

\(\displaystyle \lim_{h\to0}\frac{f(x+h)-f(x)}{h}\)

We define \(\displaystyle f(x+h)\) and \(\displaystyle f(x)\) as follows:

\(\displaystyle f(x)=2x^2+3x-4\)

\(\displaystyle \\f(x+h)=2(x+h)^2+3(x+h)-4\\ =2(x^2+2xh+h^2)+3x+3h-4 \\ =2x^2+4xh+2h^2+3x+3h-4\)

We can then define their difference:

\(\displaystyle f(x+h)-f(x)\)

\(\displaystyle =2x^2+4xh+2h^2+3x+3h-4-(2x^2+3x-4)\)

\(\displaystyle =4xh+2h^2+3h\)

Then, we divide by h to prepare to take the limit:

\(\displaystyle \frac{f(x+h)-f(x)}{h}=\frac{4xh+2h^2+3h}{h}=4x+2h+3\)

Then, the limit will give us the equation of the derivative.

\(\displaystyle f'(x)=\lim{h\to0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to0}4x+2h+3=4x+3\)

Now, we must realize that the slope of the line tangent to the curve at the given point is equivalent to the derivative at the point. So if we define our tangent line as: \(\displaystyle y-P_y=m(x-P_x)\), then this m is defined thus:

\(\displaystyle m=f'(P_x)=f'(3)=4(3)+3=15\)

Therefore, the equation of the line tangent to the curve at the given point is:

\(\displaystyle y-23=15(x-3)\implies y=15x-22\)

First, find the slope of this tangent line by taking the derivative:

\(\displaystyle y' = 6x - 2\)

Plugging in 1 for x:

\(\displaystyle 6(1)-2 = 6-2 = 4\) So the slope is 4

Now we need to find the y-coordinate when x is 1, so plug 1 in to the original equation:

\(\displaystyle y = 3(1)^2 -2(1) +1 = 3 - 2 + 1 = 2\)

To write the equation, use point-slope form and then use algebra to change to slope-intercept like the answer choices:

\(\displaystyle y - 2 = 4(x - 1 )\) distribute the 4

\(\displaystyle y - 2 = 4x - 4\) add 2 to both sides

\(\displaystyle y=4x-2\)