What is the point-slope form of a line with a slope -3 that contains the point (10,-1)

Mathematics

1 answer:

Citrus2011 [14]2 years ago

4 0

Here's the formula:

y - y1 = m(x - x1)

Substitute numbers accordingly:

y1: so 1 goes in the y1 spot (you switch the signs because it was already negative)

x1: and 10 goes to the x1 spot

m: -3 belongs in m

You might be interested in

What is the y-value of the vertex of the function f(x)=-(x-3)(x+11)

lutik1710 [3]

so, this is a quadratic equation, meaning two solutions, and we have a factored form of it, meaning you can get the solutions by simply zeroing out the f(x).

$\bf \stackrel{f(x)}{0}=-(x-3)(x+11)\implies 0=(x-3)(x+11)\implies x= \begin{cases} 3\\ -11 \end{cases} \\\\\\ \boxed{-11}\stackrel{\textit{\large 7 units}}{\rule[0.35em]{10em}{0.25pt}}-4\stackrel{\textit{\large 7 units}}{\rule[0.35em]{10em}{0.25pt}}\boxed{3}$

so the zeros/solutions are at x = 3 and x = -11, now, bearing in mind the vertex will be half-way between those two, checking the number line, that midpoint will be at x = -4, so the vertex is right there, well, what's f(x) when x = -4?

$\bf f(-4)=-(-4-3)(-4+11)\implies f(-4)=7(7)\implies f(-4)=49 \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill \stackrel{vertex}{(-4~~,~~49)}~\hfill$