what’s the equation in slope intercept form

To find the equation of the tangent line in slope-intercept form, we need two pieces of information: the slope of the tangent line and the coordinates of the point where the tangent line touches the curve.

We already have the slope of the tangent line at \( x = 1 \), which is \(-8\).

Next, we need the coordinates of the point on the curve where \( x = 1 \). We can find the \( y \)-coordinate by substituting \( x = 1 \) into the original function:

\[ y = (1)^4 - 4(1)^3 + 2 \]
\[ y = 1 - 4 + 2 \]
\[ y = -1 \]

So, the point of tangency is \((1, -1)\).

Now we can use the point-slope form of the equation of a line:

\[ y - y_1 = m(x - x_1) \]

where \((x_1, y_1)\) is the point of tangency \((1, -1)\) and \( m \) is the slope \(-8\).

\[ y - (-1) = -8(x - 1) \]
\[ y + 1 = -8(x - 1) \]
\[ y + 1 = -8x + 8 \]
\[ y = -8x + 8 - 1 \]
\[ y = -8x + 7 \]

Therefore, the equation of the tangent line in slope-intercept form is:

\[ y = -8x + 7 \]

Would you like further details or have any questions?

Here are 5 related questions you might want to ask:

1. How do you convert from point-slope form to slope-intercept form?
2. What is the significance of the y-intercept in the equation of a line?
3. Can you explain how to find the equation of a tangent line for other types of functions?
4. How does the slope of a tangent line relate to the derivative of the function?
5. What are some applications of finding tangent lines in real-world problems?

**Tip:** When finding the equation of a tangent line, always ensure that you correctly identify the slope and the coordinates of the point of tangency.

Learn how to find the equation of a tangent line in slope-intercept form using the point-slope form. This problem covers the equation of a line, slope calculations, and application of point-slope form to find the tangent line equation at a specific point on a curve. Suitable for students in Grades 9-12.

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