To find the equation of a line tangent to a curve at a certain point, we first need to find the slope of the curve at that point. To find the slope of a function at any point, we need its derivative:

$\displaystyle y=x^2\rightarrow y'=2x$

Now we can plug in the x value of the given point to find the slope of the function, and therefore the slope of tangent line, at that point:

$\displaystyle y'=2x$

$\displaystyle y'(-1)=2(-1)=-2\rightarrow m=-2$

Now that we have our slope, we can simply plug this value in with the given point to solve for the y intercept of the tangent line:

$\displaystyle y=mx+b$

$\displaystyle 1=(-2)(-1)+b$

$\displaystyle b=-1$

We have calculated the slope of the tangent line and its y intercept, so the equation for the line tangent to the curve  $\displaystyle y=x^2$  at the point  $\displaystyle (-1,1)$  in standard form is:

$\displaystyle y=-2x-1$

First we find the slope of the tangent line by taking the derivative of the function and plugging in the $\displaystyle x$-value of the point where we want to know the slope:

$\displaystyle f(x)=3x^2+4x-8$

$\displaystyle f'(x)=6x+4$

$\displaystyle f'(2)=6(2)+4=16$

Now that we know the slope of the tangent line, we can plug it into the equation for a line along with the coordinates of the given point in order to calculate the $\displaystyle y$-intercept:

$\displaystyle y=mx+b$

$\displaystyle 12=(16)(2)+b$

$\displaystyle b=-20$

We now have $\displaystyle m$ and $\displaystyle b$, so we can write the equation of the tangent line:

$\displaystyle y=16x-20$

To find the equation of a line tangent to the curve $\displaystyle y=x^{2}-2x+1$ at the point $\displaystyle (3,4)$, we must first find the slope of the curve at the point $\displaystyle (3,4)$ by solving the derivative $\displaystyle y'=2x-2$ at that point:

$\displaystyle y'(3)=2(3)-2=6-2=4\rightarrow m=4$

Given the slope, we can now plug the given point and the slope at that point into the slope-intercept form of the tangent line $\displaystyle y=mx+b$ and solve for the $\displaystyle y$-intercept $\displaystyle b$:

$\displaystyle (4)=(4)(3)+b$

$\displaystyle 4=12+b$

$\displaystyle -8=b$

Given our slope, the chosen point, and the $\displaystyle y$-intercept, we have the equation of our tangent line:

$\displaystyle y=4x-8$

To find the equation of a line tangent to the curve $\displaystyle y=2x^{2}+5x+2$ at the point $\displaystyle (2,-1)$, we must first find the slope of the curve at the point $\displaystyle (2,-1)$ by solving the derivative $\displaystyle y'=4x+5$ at that point:

$\displaystyle y'(2)=4(2)+5=8+5=13\rightarrow m=13$

Given the slope, we can now plug the given point and the slope at that point into the slope-intercept form of the tangent line $\displaystyle y=mx+b$ and solve for the $\displaystyle y$-intercept $\displaystyle b$:

$\displaystyle (-1)=(13)(2)+b$

$\displaystyle -1=26+b$

$\displaystyle -27=b$

Given our slope, the chosen point, and the $\displaystyle y$-intercept, we have the equation of our tangent line:

$\displaystyle y=13x-27$

First find the slope of the tangent line by taking the derivative of the function and plugging in the x value of the given point to find the slope of the curve at that location:

$\displaystyle f(x)=-2x^2+9x-17$

$\displaystyle f'(x)=-4x+9$

$\displaystyle f'(2)=-4(2)+9=1$

So the slope of the tangent line to the curve at the given point is  $\displaystyle m=1$.  The next step is to plug this slope into the formula for a line, along with the coordinates of the given point, to solve for the value of the y intercept of the tangent line:

$\displaystyle y=mx+b$

$\displaystyle -7=1(2)+b\rightarrow b=-9$

We now know the slope and y intercept of the tangent line, so we can write its equation as follows:

$\displaystyle y=x-9$

To find the equation of a line tangent to the curve $\displaystyle y=4x^{2}-3x+7$ at the point $\displaystyle (-2,-3)$, we must first find the slope of the curve at the point $\displaystyle (-2,-3)$ by solving the derivative $\displaystyle y'=8x-3$ at that point:

$\displaystyle y'(-2)=8(-2)-3=-16-3=-19\rightarrow m=-19$

Given the slope, we can now plug the given point and the slope at that point into the slope-intercept form of the tangent line $\displaystyle y=mx+b$ and solve for the $\displaystyle y$-intercept $\displaystyle b$:

$\displaystyle (-3)=(-19)(-2)+b$

$\displaystyle -3=38+b$

$\displaystyle -41=b$

Given our slope, the chosen point, and the $\displaystyle y$-intercept, we have the equation of our tangent line:

$\displaystyle y=-19x-41$

f(x) in this case gives us a parabola. If we factor the f(x) we get:

$\displaystyle \small f(x)=(x-3)(x-3)$

This equation has one zero and that is at x=3. This also means we have a minimum at x=3.

The slope of a line tangent to a minimum or maximum is always 0, so our slope is zero.

The slope of the tangent line to a curve at any point is simply the slope of the curve at that point. To find the slope of the function at any point, we take the derivative:

$\displaystyle f(x)=-2x^2+13x-7$

$\displaystyle f'(x)=-4x+13$

Now we can plug in the x value of the given point, $\displaystyle (1,4)$, which gives us the slope of the tangent line to the curve at that point:

$\displaystyle f'(x)=-4x+13$

$\displaystyle f'(1)=-4(1)+13=9$

To calculate the slope of a tangent line at a particular point, we need to know the slope of the curve at that point. In order to find the slope of a curve at any point, we need to calculate its derivative:

$\displaystyle y=3x^2-5x+7$

$\displaystyle y'=6x-5$

The derivative describes the slope of the curve at any point, so we need to plug in the $\displaystyle x$ value of the given point  $\displaystyle (2,15)$  to find the slope of the tangent line at that location:

$\displaystyle y'=6x-5$

$\displaystyle y'(2)=6(2)-5=7$

First find the derivative of the function, and then plug in the given x coordinate, which will give the slope of the function at that location. The slope of the tangent line to a curve at a given location is equal to the slope of the function at that location:

$\displaystyle f(x)=6x^5-7x^4+2x^3$

$\displaystyle f'(x)=30x^4-28x^3+6x^2$

$\displaystyle f'(1)=30(1)^4-28(1)^3+6(1)^2=8$