SOLUTION: I need help! 1. write the equation of a line whose slope is -2 and passes through (-1,6). Write answer in slope-intercept form. 2. Solve for x: (2x-7)(x+4)=0 3. find th

Algebra ->  Functions -> SOLUTION: I need help! 1. write the equation of a line whose slope is -2 and passes through (-1,6). Write answer in slope-intercept form. 2. Solve for x: (2x-7)(x+4)=0 3. find th      Log On


   



Question 147711: I need help!
1. write the equation of a line whose slope is -2 and passes through (-1,6). Write answer in slope-intercept form.
2. Solve for x: (2x-7)(x+4)=0
3. find the vertex of f(x)=3x^2-6x+8
4. find the y-intercept of the cubic f(x)=(x-2)(x+1)(x-3)

Found 2 solutions by stanbon, mangopeeler07:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
1. write the equation of a line whose slope is -2 and passes through (-1,6). Write answer in slope-intercept form.
y = mx + b
y = 6 when x = -1, and m= -2 ; solve for "b":
6 = -2*-1 + b
b = 4
EQUATION:
y = -2x + 4
-------------------------
2. Solve for x: (2x-7)(x+4)=0
x = 7/2 or x = -4
-------------------------
3. find the vertex of f(x)=3x^2-6x+8
Vertex occurs when x = -b/2a = 6/6 = 1
f(1) = 3-6+8 = 5
Vertex: (1,5)
-------------------------
4. find the y-intercept of the cubic f(x)=(x-2)(x+1)(x-3)
Let x = 0.
f(0) = -2*1*-3 = 6
======================
Cheers,
Stan H.

Answer by mangopeeler07(462) About Me  (Show Source):
You can put this solution on YOUR website!
1. y=mx+b where m is the slope and b is the y-intercept (what y is when x is zero). So y=-2x+b. To find b, you know that the slope is rise/run or -2/1. So take the point (-1,6) and add 1 to x and subtract 2 from y. You get 0%2C4. So the y-intercept is 4. So y=-2x%2B4.

2. (2x-7)(x+4)=0. Just take each expression separately and set each one equal to zero. what minus seven equals zero? Seven. So you know 2x=7. So x there is 7/2. What plus four equals zero? x=-4 there. So the solutions for this equation are x=7/2;-4.

3.find the vertex of f(x)=3x^2-6x+8. vertex: the value of y that does not repeat. To get this, plug in a few consecutive values of x and see which one does not repeat an answer. That would be the vertex. In this case, I will let you know that it is at (1,5). Because:
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 3x%5E2%2B-6x%2B8+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%28-6%29%5E2-4%2A3%2A8=-60.

The discriminant -60 is less than zero. That means that there are no solutions among real numbers.

If you are a student of advanced school algebra and are aware about imaginary numbers, read on.


In the field of imaginary numbers, the square root of -60 is + or - sqrt%28+60%29+=+7.74596669241483.

The solution is

Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+3%2Ax%5E2%2B-6%2Ax%2B8+%29


5 is the only y value that does not repeat.

4. find the y-intercept of the cubic f(x)=(x-2)(x+1)(x-3). y=f(x), so that means find f(0), because the y-intercept is what y is when x is 0. So plug in 0 and get f%280%29=%280-2%29%280%2B1%29%280-3%29. Or f%280%29=%28-2%29%281%29%28-3%29. Multiply it all out and get 6. So the y-intercept is 6.