How do you find the tangent line of
#f(x) = 3-2x^2 # at x=-1?

Question

Christopher Allers · Accepted Answer

Take the derivative and plug in the desired #x=-1# to get the slope. Then back-solve for the y-intercept. You will find the expression is:

The slope-intercept equation for a given line is #y=mx+b#. We know by definition that a tangent line has the same value as the original function at the point of tangency, and the slope is the same as the functions' at that point. We can calculate the slope from the derivative of the function: #f(x)=3-2x^2# #(dy)/(dx)f(x)=f'(x)=-4x# #f'(-1)=-4(-1)=color(blue)(4)# Now that we have our slope, we can evaluate the original function and solve for the y-intercept: #f(-1)=3-2(-1)^2# #f(-1)=3-2# #f(-1)=1# #f(-1)=4(-1)+b# #1=-4+b# #color(blue)(b=5)# Now, we know the slope and intercept, and can write the equation for the tangent line: #color(green)(y=4x+5)#

Emma Aldridge · Answer

undefined To find the tangent line of f(x) = 3-2x^2 at x=-1, we need to find the derivative of the function and evaluate it at x=-1. The derivative of f(x) is f'(x) = -4x. Evaluating f'(-1), we get f'(-1) = -4(-1) = 4. Therefore, the slope of the tangent line at x=-1 is 4. To find the equation of the tangent line, we use the point-slope form y - y1 = m(x - x1), where (x1, y1) is the point on the curve. Plugging in x=-1, y=f(-1), and m=4, we have y - (3-2(-1)^2) = 4(x - (-1)). Simplifying this equation gives us the equation of the tangent line as y = 4x + 7.

Ezra Adams · Answer

undefined To find the tangent line of $ f(x) = 3 - 2x^2 $ at $ x = -1 $, follow these steps:

1. Find the derivative of $ f(x) $ using the power rule: $ f'(x) = -4x $.
2. Evaluate the derivative at $ x = -1 $: $ f'(-1) = -4(-1) = 4 $.
3. Use the point-slope form of a line, $ y - y_1 = m(x - x_1) $, where $ m $ is the slope and $ (x_1, y_1) $ is a point on the line.
4. Substitute the values $ x_1 = -1 $ and $ m = 4 $ into the point-slope form.
5. Simplify the equation to obtain the tangent line equation.

Therefore, the tangent line of $ f(x) = 3 - 2x^2 $ at $ x = -1 $ is $ y = 7 - 4x $.

Evelyn Agard · Answer

undefined To find the tangent line of $f(x) = 3 - 2x^2$ at $x = -1$, follow these steps:

1. Find the derivative of $f(x)$ with respect to $x$ to get the slope of the tangent line.
2. Evaluate the derivative at $x = -1$ to find the slope of the tangent line at that point.
3. Use the point-slope form of the equation of a line, $y - y_1 = m(x - x_1)$, where $m$ is the slope of the tangent line and $(x_1, y_1)$ is the point of tangency, which is $(-1, f(-1))$.
4. Plug in the values of $m$ and $(x_1, y_1)$ into the point-slope form to get the equation of the tangent line.

Let's go through the steps:

1. Find the derivative of $f(x)$:
$$f'(x) = \frac{d}{dx}(3 - 2x^2) = -4x$$

2. Evaluate the derivative at $x = -1$:
$$f'(-1) = -4(-1) = 4$$

3. Determine the point of tangency:
$$f(-1) = 3 - 2(-1)^2 = 3 - 2 = 1$$
So, the point of tangency is $(-1, 1)$.

4. Use the point-slope form:
$$y - 1 = 4(x + 1)$$

5. Simplify the equation:
$$y - 1 = 4x + 4$$
$$y = 4x + 5$$

Therefore, the equation of the tangent line to $f(x) = 3 - 2x^2$ at $x = -1$ is $y = 4x + 5$.

How do you find the tangent line of #f(x) = 3-2x^2 # at x=-1?