Find all of the indicated values for each quadratic equation. 1. f(x)=-2x²-2x-5 2. y=x²+4x+2 3. f(x)=3x²-6x+1 4. y=-4x²+8x-8

Question

Help me! Thank you! 

Answer 1

You have still not stated what actual values you are looking for.

Given  f(x) = ax²+ bx + c

If you need the x-intercepts (where f(x) = 0) you can use the quadratic

equation on the problems.   If you need the vertex of the parabola in

question it is located at (-b/2a, f(-b/2a)).

 

Let's look at your first problem:

f(x) = -2x²-2x-5

a = -2, b = -2, c= -5

 

This is a parabola opening downward since the coefficient of the

squared term is negative, so the vertex will be the maximum

point of the graph.  The vertex is located at (-b/2a, f(-b/2a))

 

-b/2a = -(-2)/2(-2) = 2/(-4) = -1/2

f(-b/2a) = f(-1/2) = -2(-1/2)² -2(-1/2)-5 = -9/2 or -4 1/2

The vertex is located at (-1/2, -4 1/2)

 

Given that the vertex is below the x axis and the parabola

opens downward, there will be no real numbers for x-intercepts.

This means we should expect the answer to the quadratic

equation -2x²-2x-5 = 0 to have "i" in the answer since the

x-intercepts will be imaginary.

 

-2x²-2x-5 = 0     a=-2, b=-2, c=-5

x = [-b ±√(b²-4ac)]/2a

 = [-(-2) ±√((-2)²-4(-2)(-5))]/(2(-2)) = [2±√(-36)]/(-4)

 

= (2±6i)/(-4) = (-1±3i)/2

 

As expected the x-intercepts at  (-1±3i)/2 are imaginary, but to the

left and right of the x coordinate of the vertex as they should be.

 

Again, we don't really know what other information you

may be looking for as "y-values" are infinite since you can

put in any value for x and compute a corresponding value

for f(x) or y

Answer 2

y=-2x²-2x-5

x = 0

y = -5

 

(If you're solving for y.  The equation is otherwise irreducible.)

 

y=x²+4x+2

x = 0

y = 2

 

3x²-6x+1

x = 0

y = 1

 

y=-4x²+8x-8

x = 0

y = -8

 

Is this what you're looking for?  I'm not sure if this addresses the question.