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How do you write an equation in slope-intercept form for the line with the given slope that contains the given point :
slope = -4; (1,-3)
the equation for it in slope intercept form (y=mx+b) is y=-4x+1 because you plug in your x and why points (1 and -3) and the slope (-4) into the equation. you get -3= (-4)(1)+b and solve for b. You'll get 1 as your b so your equation in slope intercept form is y=-4x+1.
For point slope form, the equation is (y-y1)=m(x-x1)...this is easier. All you have to do is plug in the numbers... your y1 is your why coordinate in (1, -3) and the x1 is your x coordinate. So...your equation would be (y+3)=-4(x-1)...get it?
Algebra and trig is my forte. I love it!
y = mx + b is the slope-intercept form. m is the slope and b is the y-intercept.
Given the slope and a point, you'll first want to plug in the values and solve for b, then write the equation.
-3 = -4(1) + b b = -3 + 4 which is 1
** The final equation is y = -4x + 1 **
y=mx+b Do it in point slope form first y-y1= m(x-x1) y - -3= -4(x-1) distribute *y+3= -4x +4 subtract frm each side y= -4x +1
Theres your answer * the y- -3 changes to y+3 because two negatives equal a positive
y=mx+b -3=-4x+1
check: -3=-4+1 -3= -4+1=-3 -3=-3
I think thats how it goess. hope it helpp any other problems with math, cmment me/message me and ill help ya(=
y - mx+b m - Slope b - why Intercept y & x - Variables
y=-4x-3
it is -3 because if you add a negative it is like minusing a positive
Point-slope form is: y-y1=m(x-x1)
So, it would be y+3=-4(x-1) and that simplifies to y+3=-4x+1 Then subtract 3. y=-4x-2
