. Given the following function, f(x)=-x^2 -8x find:
(a) vertex, (b) axis of symmetry, (c) intercepts, (d) domain, (e) range,
(f) intervals where the function is increasing,
(g) intervals where the function is decreasing, and
(h) the graph of the function. Please show all of your work.
(Points : 15)
vertex: x = -b/2a = 8/-2 = -4, so y=-48: (-4,16)
axis: x = -4
x-intercepts: 0,-8
y-intercepts: 0
domain: all real x
range: all real y <= 16
y increasing where y' > 0
y' = -2x-8
y'>0 when x < -4
y decreasing for x > -4
To find the information requested about the function f(x)=-x^2 -8x, let's go through each step:
(a) To find the vertex of the function, we first need to remember the formula for the x-coordinate of the vertex: x = -b / 2a. In our function, a = -1 and b = -8. Plugging these values into the formula, we get:
x = -(-8) / (2*(-1)) = 8 / (-2) = -4.
To find the corresponding y-coordinate of the vertex, we substitute the value of x back into the function:
f(-4) = -(-4)^2 - 8(-4) = -16 + 32 = 16.
So, the vertex of the function is (-4, 16).
(b) The axis of symmetry is simply the x-coordinate of the vertex, which is x = -4.
(c) To find the x-intercepts, we set f(x) equal to 0 and solve for x:
0 = -x^2 - 8x.
To solve this quadratic equation, we can try factoring:
0 = -x(x + 8).
Setting each factor equal to zero, we get x = 0 and x = -8.
Therefore, the x-intercepts are (0, 0) and (-8, 0).
To find the y-intercepts, we substitute x = 0 into the function:
f(0) = -(0)^2 - 8(0) = 0.
So, the y-intercept is (0, 0).
(d) The domain is the set of all possible x-values for which the function is defined. In this case, there are no restrictions on the domain, so the domain is all real numbers.
(e) The range is the set of all possible y-values that the function can take. Since the coefficient of x^2 is -1 and negative, the maximum value occurs at the vertex and the function decreases indefinitely from there. Therefore, the range is (-∞, 16]. The ']' indicates that 16 is included in the range.
(f) To determine the intervals where the function is increasing, we need to find where the derivative of the function is positive. Taking the derivative of f(x) with respect to x, we have:
f'(x) = -2x - 8.
Setting f'(x) > 0, we solve for x:
-2x - 8 > 0.
Adding 8 to both sides:
-2x > 8.
Then, dividing by -2 (and note the change in inequality direction since we divide by a negative number):
x < -4.
So, the function is increasing for all x < -4.
(g) To determine the intervals where the function is decreasing, we need to find where the derivative of the function is negative. Taking the derivative of f(x) with respect to x, we have:
f'(x) = -2x - 8.
Setting f'(x) < 0, we solve for x:
-2x - 8 < 0.
Adding 8 to both sides:
-2x < 8.
Then, dividing by -2 (and note the change in inequality direction since we divide by a negative number):
x > -4.
So, the function is decreasing for all x > -4.
(h) To graph the function, we start by plotting the vertex (-4, 16) and the intercepts (0, 0) and (-8, 0). Since we know the graph is a downward-facing parabola, it will open downwards.
Additionally, since the axis of symmetry is x = -4, we can use this information to determine other points on the graph by reflecting points across the axis of symmetry. For example, if we have a point (2, y) on one side of the axis of symmetry, we know there will be another point (-10, y) at the same height on the other side of the axis of symmetry.
By connecting the points and adding more if desired, we can sketch the graph of the function f(x) = -x^2 - 8x.
I hope this explanation helps! Let me know if you have any further questions.