For what values of k will the function y=6x^2-8x+k have one zero: Two zeros: No zeros

Step-by-step explanation:

consider 6x^2-8x +k

disc .=(-8)^2-4*6*k=64-24k

it has no solution if 64-2k<0

or 64<2k

or 2k>64

k>32

it has two zeros if 64-2k≥0

or 64≥ 2k

or 2k≤64

or k≤ 32

it has one zero if Disc. =0

64-2k=0

2k=64

k=32

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step-by-step explanation:

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-3

step-by-step explanation:

a drawing really does here:

so i draw the horizontal line y=2 and the vertical line x=-2 at this intersection i put b (this is where my right angle is going to be).   so point b is at (-2,2).   point a is 5 units long and is also on y=2 so this point has to be (3,2) or (-7,2) -doesn't matter i will go with (3,2).

now we know c is at (-2,y) we don't know the y yet bet we know that c is on x=-2.

we have the area of the triangle is .5bh=12.5

we know the base length is 5 (ab)

we have the height is (2-y) if you look at your

plug in

.5(5)(2-y)=12.5

solve for y to get the y-coordinate for c

2.5(2-y)=12.5

divide both sides by 2.5 giving:

2-y=5

now subtract 2 on both sides

-y=3

y=-3