How do you find the range of the equation #y = -x^2 – 6x – 13#?

Answer 1

Eva Adamson

Range of #y = [-4,-oo)#

#y = -x^2-6x-13#

#y# is a quadratic function, represented on the #xy-#plane as a parabola of the form: #ax^2+bx+c#

The vertex of the parabola will be at #x=( -b)/(2a)#

In our case, #b=-6, a=-1#

Hence, #x_(vertex) = (6)/(-2) =-3#

Since #a<0# then #y(x_(vertex) )# will be a maximum of #y#

#:. y_max = y(-3) = -(-3)^2+6*3-13 = -9+18-13=-4#

#:. # the greatest value of #y# is #-4#

Since #y# has no lower bounds, the range of #y# is #[-4, -oo)#

As can be seen from the graph of #y# below.

graph{-x^2-6x-13 [-23.18, 22.45, -15.1, 7.71]}

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Answer 2

Nathan Axel

To find the range of the equation $y = -x^2 - 6x - 13$ , we need to determine the maximum or minimum value of the quadratic function. Since the coefficient of $x^2$ is negative, the parabola opens downwards, indicating that the maximum value occurs at the vertex. To find the vertex, we use the formula $x = -\frac{b}{2a}$ , where $a$ is the coefficient of $x^2$ (in this case, $a = -1$ ) and $b$ is the coefficient of $x$ (in this case, $b = -6$ ). Substituting these values into the formula, we find $x = -\frac{-6}{2(-1)} = 3$ . Next, we substitute $x = 3$ into the equation $y = -x^2 - 6x - 13$ to find the corresponding $y$ -coordinate: $y = -(3)^2 - 6(3) - 13 = -9$ . Therefore, the vertex of the parabola is $(3, -9)$ . Since the parabola opens downwards, the range of the equation is $y \leq -9$ .

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.