What is the range of the following parabola? f(x) = -6x^2 - 10x + 13?

Question

I know that the domain of all parabola's is (-infinity, infinity), and I know to find the range I need to convert the equations from standard form to vertex form. However, when I enter the answer I get from doing that, it says it is the wrong answer. (I am using WebWork) Anyone have any suggestions?

Thanks alot for our help!

Answer 1

Yeah we have to put it into vertex form:

-6(x^2+(5/3)x)+13

-6(x^2+(5/3)x+(25/36))+13+(25/36)*6

-6(x+5/6)^2+13+(25/6)

-6(x+5/6)^2+(78/6)+(25/6)

-6(x+5/6)^2+(103/6)

Vertex=((-5/6),(103/6))

So since it opens downward, it's highest point is 103/6.

So its range is (-oo, 103/6)

Answer 2

f(x) = -6x^2 - 10x + 13

the vertex :

V(-b/2a ; -Δ/4a)

-b/2a=10/(-12)= - 5/6

-Δ/4a= -(b^2-4ac)/4a= -(100+312)/(-24)=412/24= 103/6  (17.16)

the parabola is oriented down (to a<0), for which the maximum value that the function (y) is reached 103/6.

the range of the function is:

[-∞; 103/6)

Answer 3

You're on the right track.

The vertex is the lowest point of this parabola.

So the range is [k, infinity] where k is the y-coordinate of the vertex.

Don't use WebWork.

Answer 4

There are many ways to determine the vertex of this parabola.  Putting the equation in vertex form may not be the easiest.  From standard form, the axis of symmetry is given by the equation:  x = -b/2a

You can then substitute this value to determine the coordinates of the vertex.

x = -b/2a = 10/12 = 5/6

Substitute 5/6, then

f (5/6) = 53/6  [you do the arithmetic]

So, the vertex is at: (5/6, 53/6)

The range would be y greater than or equal to 53/6