The line x = k intersects the graph of the parabola x = -2y^2 - 3y + 5 at exactly one point. What is k? Thank you! :)

49/8 is the value of k

Step-by-step explanation:

We have the system

x = -2y^2 - 3y + 5

x=k

We want to find k such that the system intersects once.

If we substitute the second into the first giving us k=-2y^2-3y+5 we should see we have a quadratic equation in terms of variable y.

This equation has one solution when it's discriminant is 0.

Let's first rewrite the equation in standard form.

Subtracting k on both sides gives

0=-2y^2-3y+5-k

The discriminant can be found by evaluating

b^2-4ac.

Upon comparing 0=-2y^2-3y+5-k to 0=ax^2+bx+c, we see that

a=-2, b=-3, and c=5-k.

So we want to solve the following equation for k:

(-3)^2-4(-2)(5-k)=0

9+8(5-k)=0

Distribute:

9+40-8k=0

49-8k=0

Add 8k on both sides:

49=8k

Divide both sides by 8"

49/8=k

-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