Question

write an equation for a line parallel to $y = -4x + 6$ and passes through the point $(-13, 63)$.

step 1:
identify the slope of the new equation.
$m = \boxed{-4}$

part 2 of 3

step 2:
substitute $m$, $x_1$, and $y_1$ into point - slope form.
$y - y1 = m(x - x1)$
$y - \boxed{63} = \boxed{-4}(x - \boxed{12})$ (the 12 is marked wrong)

part 3 of 3

step 3:
solve for $y$. your equation should end up in slope - intercept form.
$y = mx + b$
\boxed{}

Explanation:

Step1: Correct \( x_1 \) in point - slope form

The point is \((- 13,63)\), so \(x_1=-13\). The point - slope form is \(y - y_1=m(x - x_1)\). Substituting \(y_1 = 63\), \(m=-4\) and \(x_1=-13\) into the formula, we get \(y - 63=-4(x-(-13))\), which simplifies to \(y - 63=-4(x + 13)\).

Step2: Expand and solve for \(y\)

First, expand the right - hand side: \(y-63=-4x-52\). Then, add 63 to both sides of the equation: \(y=-4x-52 + 63\).

Step3: Simplify the equation

Simplify the constant terms: \(-52 + 63 = 11\). So the equation of the line in slope - intercept form is \(y=-4x + 11\).

Answer:

\(y=-4x + 11\)