Tangent Lines - GMAT Quantitative

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

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:

Now we can plug in the x value of the given point, , which gives us the slope of the tangent line to the curve at that point:

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:

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:

First find the derivative of the function. The slope of the tangent line to a curve at a given point is equal to the slope of the function at that point, and the derivative of a function tells us its slope at any point. Plugging the given point into the equation for the derivative, we can calculate the slope of the function, and therefore the slope of the tangent line, at that point:

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:

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:

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:

We have calculated the slope of the tangent line and its y intercept, so the equation for the line tangent to the curve at the point in standard form is:

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

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 -intercept:

We now have and , so we can write the equation of the tangent line:

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:

So the slope of the tangent line to the curve at the given point is . 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:

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

To find the equation of a line tangent to the curve at the point , we must first find the slope of the curve at the point by solving the derivative at that point:

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 and solve for the -intercept :

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

To find the equation of a line tangent to the curve at the point , we must first find the slope of the curve at the point by solving the derivative at that point:

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 and solve for the -intercept :

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

To find the equation of a line tangent to the curve at the point , we must first find the slope of the curve at the point by solving the derivative at that point:

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 and solve for the -intercept :

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

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:

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:

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:

We have calculated the slope of the tangent line and its y intercept, so the equation for the line tangent to the curve at the point in standard form is:

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

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 -intercept:

We now have and , so we can write the equation of the tangent line:

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:

So the slope of the tangent line to the curve at the given point is . 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:

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

To find the equation of a line tangent to the curve at the point , we must first find the slope of the curve at the point by solving the derivative at that point:

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 and solve for the -intercept :

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

To find the equation of a line tangent to the curve at the point , we must first find the slope of the curve at the point by solving the derivative at that point:

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 and solve for the -intercept :

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

To find the equation of a line tangent to the curve at the point , we must first find the slope of the curve at the point by solving the derivative at that point:

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 and solve for the -intercept :

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

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:

The derivative describes the slope of the curve at any point, so we need to plug in the value of the given point to find the slope of the tangent line at that location:

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

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.