Answer :
Answer:
[tex]\fbox{y = -3x + 5}[/tex]
Step-by-step explanation:
You are given the slope and the y-intercept of the line; so you can substitute these values into slope-intercept form: [tex]y=mx+b[/tex];
- where [tex]m= \ $slope = -3[/tex]
- and [tex]y-$int = 5[/tex]
Plugging these values into slope-intercept form gives:
- [tex]y=(-3)x+(5)[/tex]
- [tex]y=-3x+5[/tex]
Another way to find the slope-intercept form of a line given the slope and a point:
We are given the slope and a point that the line passes through, so we can use the point-slope equation to find the slope-intercept form of the line.
The point that the line passes through is the y-intercept: [tex](0,5)[/tex].
Point-slope form:
- [tex]y-y_1=m(x-x_1)[/tex]
where [tex](x_1, \ y_1)[/tex] are the coordinates of the point that the line passes through and [tex]m= $ slope of the line.[/tex]
Substitute [tex]m=-3[/tex] and [tex](0,5)[/tex] into the point-slope form equation.
- [tex]y-(5)=-3(x-(0))[/tex]
Simplify the equation on both sides.
- [tex]y-5=-3(x)[/tex]
Add 5 to both sides of the equation.
- [tex]y=-3x+5[/tex]
This is in slope-intercept form: [tex]y=mx+b[/tex], so we are done.
The answer is [tex]D) \ $y = -3x+5[/tex].