Just to be clear, the two points are (-2,2) and (0,5). The slope (rise/run) is (5-2)/(0 - -2) = 3/2. So y = (3/2)*x + b. Put one of the point into this and solve for b: 5 = (3/2)*0 + b; b = 5. The equation is y = (3/2)*x + 5. Do a check that the other point (-2,2) satisfies this equation {and it does}.
We know that its slope is negative, but without an equation or some points the line passes through we can't determine the actual value of the slope.
If you mean points of (2, -2) and (-4, 22) then the equation is y = -4x+6
The equation for the given points is y = x+4 in slope intercept form
It is y = 2.
Answer this question…y = 2x + 6
Write the equation of the line that passes through the points (3, -5) and (-4, -5)
We know that its slope is negative, but without an equation or some points the line passes through we can't determine the actual value of the slope.
If you mean points of (2, -2) and (-4, 22) then the equation is y = -4x+6
The equation for the given points is y = x+4 in slope intercept form
It is y = 2.
Y= -3x + 8
Plug both points into the equation of a line, y =m*x + b and then solve the system of equations for m and b to get equation of the line through the points.
In order to find the equation of a tangent line you must take the derivative of the original equation and then find the points that it passes through.
If you mean points of (-4, 2) and (4, -2) Then the straight line equation works out as 2y = -x
Points: (0. 5) and (2, 3) Slope: -1 Equation: y = -x+5
Answer this question…y = 2x + 6
If the points are (1,5) and (0,0) y = 5x