Choose the slope-intercept equation of the line that passes through the point shown (with graph)?

You are correct.

m12172012

The standard conference for a straight line graph is y = mx + c, where m is the gradient. This also signifies that the coefficient of x is the gradient The gradient is a evaluation between points so solving this equation for m (I will not do it right here, simply believe me...:-)) m = (y1 - y2) divided by using (x1 - x2) => m = (4 - -6) divided through (-1 -four) => m = 10 divided through -5 => m = -2 which means that your line will seem like a back shrink, as the gradient is lower than zero (ii) the common expression for a straight line is y = mx + c but we've got cleverly decided that m = -2 (from above) So, y = -2x + c We even have coordinates for 2 features; let's recall the primary one (-1,four) we all know in this case that y = 4, and x = -1 for that reason, substituting the phrases that we all know, will support us solve for c 4 = (-2 x -1) + c => 4 = 2 + c => 4-2 = c consequently c = 2 for this reason the equation( y = mx +c) is y = -2x + 2 (iii) the x intercept is the point where y = 0 accordingly, headquartered on our equation above 0 = -2x + 2 => 2x = 2 => x = 1 Intercept is a (1,zero) (iv) For (2,-2) to be on the road, requires that we satisfy the equation for the road y = -2x + 2 If x = 2, then y = (-2 extended via 2) + 2 => y = -4 + 2 => y = -2 So, if x = 2, y have to equal -2 to lie on the road. This factor (2,-2) satisfies the equation for the straight line, so (2,-2) is a point on the line (v) : 1) that the x intercept occurs at (1,zero) 2) the gradient is -2 To be parallel, the new line have got to have the identical gradient. Lets make x = 2 ( for illustration; you can also decide on any price for x) thus, the x intercept for the brand new line could be at (2,zero), as y would = zero at this factor (as above) y = mx + c => 0 = (-2 x 2) + c => four = c therefore, a line parallel to y = -2x + 2 is y = -2x + four good luck!

I get y = (-1/2)x - 7/2

The given graph shows a straight line that has a slope of +2 and a y-intercept of 2

From y = mx+b, its equation is

y = 2x + 2

The line perpendicular to y = 2x + 2 will have a slope of -1/2, since perpendicular lines have slopes that are negative reciprocals of one another. The negative reciprocal of 2 is - 1/2

So its equation is y = (-1/2)x + b

The line passes through (-5,-1)

Substituting -1 for y and -5 for x gives

-1 = (-1/2)(-5) + b

-1 = 5/2 + b

-2/2 -5/2 = b

-7/2 = b

The required equation is y = (-1/2)x -7/2, and this solution is not in your list.

That's my story and I'm sticking to it !!