Mr. Shaw graphs the function f(x) = –5x + 2 for his class. The line contains the point (-2, 12). What is the point-slope form of the equation of the line he graphed?

y – 12 = –5(x + 2)
y – 12 = 2(x + 2)
y + 12 = 2(x – 2)
y + 12 = –5(x – 2)

Respuesta :

f(x) = -5x + 2...this can be changed to y = -5x + 2...so ur slope(m) = -5

point-slope form : y - y1 = m(x - x1)
slope(m) = -5
(-2,12)...x1 = -2 and y1 = 12
now we sub
y - 12 = -5(x -(-2) =
y - 12 = -5(x + 2) <====

The point-slope form: [tex]y-y_0=m(x-x_0)[/tex]

The slope0intercept form: [tex]y=mx+b[/tex]

[tex]f(x)=-5x+2\Rightarrow m=-5[/tex]

[tex](-2;\ 12)\Rightarrow x_0=-2;\ y_0=12[/tex]

subtitute:

[tex]y-12=-5[x-(-2)]\to\boxed{y-12=-5(x+2)}[/tex]