Can someone help me and verify if the answer I picked is correct?
1. Which function does not have a maximum value?
a) x^2 + 9
b) g(x) = 9 - x - x^2
c) h(x) = -(x+1)^2
d) k(x) = -4 + x - x^2
The answer is A because letter b,c,d opens the parabola downwards?
Myra is riding a Ferris Wheel. Her height h(t), in metres above the ground at time t seconds, can be modeled by
h(t) = 10sin(6(t - 20°)) + 10. At what time will Myra's car be at its greatest height?
a) t=20 s
b) t=35 s
c) t=40 s
d) t=55 s
Is it C?
3. At which point on the graph of f(x)= -x^2 - 2x +15 is the slope of the tangent 0?
a) (-2, 15)
b) (-1,16)
c) (0,15)
d) (3,0)
Help~
1 and 3 are correct
2. First of all do not put a ° symbol behind the 20, that makes no sense,
10 sin(6(t-20)) + 10 has a maximimum value of 20 when sin(6(t-20)) is 1
we know sinπ/2 = 1, so 6(t-20) = π/2
t-20 = π/12
t = 20 + π/12 = 20.26
Whoever made up this equation does not seem to understand that if t is defined in seconds, then you cannot use degrees in the equation, yet if t=35
then we get 10sin(6(35-20)+10 = 10(sin 90) + 10 = 20
But the 90 would have to be degrees and the 35 was in seconds. What a mess-up!
The equation is bogus, and makes no sense.
Thanks~ I understand #2 now.
If I have more questions, do I post them here or post a new question?
1. To determine which function does not have a maximum value, we need to look at the vertex of each function. The vertex is the point on the parabola where the function reaches its maximum or minimum value.
a) The function x^2 + 9 is a quadratic function in the form of y = ax^2 + bx + c. In this case, a = 1, b = 0, and c = 9. The vertex of a quadratic function is given by the formula x = -b/2a and y = f(x). Plugging in the values, we get x = 0 and y = 9. Therefore, the vertex is (0, 9), which means the function has a maximum value of 9.
b) The function g(x) = 9 - x - x^2 is also a quadratic function. Using the same approach as above, the vertex is (1/2, 35/4). This means the function has a maximum value.
c) The function h(x) = -(x+1)^2 is a quadratic function in vertex form. By expanding and simplifying, we get h(x) = -x^2 - 2x - 1. The vertex can be obtained using the same method as before, which gives us (-1, -1). This means the function has a maximum value.
d) The function k(x) = -4 + x - x^2 is yet another quadratic function. Its vertex is (1/2, -15/4), indicating a maximum value.
Based on the analysis above, the answer is none of the options (a), (b), (c), or (d), as all of the functions have a maximum value.
2. To find the time at which Myra's car is at its greatest height, we need to find the maximum value of the function h(t) = 10sin(6(t - 20°)) + 10. The maximum value of a sine function occurs when the angle inside the sine function is equal to 90 degrees or π/2 radians.
Setting 6(t - 20°) equal to 90 degrees, we have:
6(t - 20°) = 90°
t - 20° = 90°/6
t - 20° = 15°
t = 15° + 20°
t = 35°
Since there are 360 degrees in a full circle, we can convert 35° to seconds:
t = 35° * (1 second / 6°)
t = 35/6 seconds
Therefore, the answer is option (b) t = 35 seconds.
3. To find the point(s) on the graph of f(x) = -x^2 - 2x + 15 where the slope of the tangent is 0, we need to find the x-coordinate(s) where the derivative of the function is equal to 0.
The derivative of f(x) can be found by applying the power rule for differentiation:
f'(x) = -2x - 2
Setting f'(x) equal to 0 and solving for x:
-2x - 2 = 0
-2x = 2
x = -1
Therefore, option (b) (-1, 16) is the point on the graph where the slope of the tangent is 0.
To verify if the chosen answers are correct, let's analyze each question step by step.
1. Which function does not have a maximum value?
In this question, we need to determine which function does not have a maximum value. The maximum value of a function occurs at the vertex of its graph. To find the vertex, we can use the formula x = -b / (2a), where a and b are the coefficients of the quadratic equation.
a) x^2 + 9: The coefficient of x^2 is 1, and the coefficient of x is 0. Plugging these coefficients into the formula, we get x = -0 / (2*1) = 0. The vertex is (0, 9), and the function has a maximum value of 9.
b) g(x) = 9 - x - x^2: The coefficient of x^2 is -1, the coefficient of x is -1, and the constant term is 9. Applying the formula, we get x = -(-1) / (2*(-1)) = 0.5. The vertex is (0.5, 8.75), and the function has a maximum value.
c) h(x) = -(x+1)^2: The coefficient of x^2 is -1, the coefficient of x is -2, and the constant term is 15. Using the formula, we find x = -(-2) / (2*(-1)) = -1. The vertex is (-1, 14), and the function has a maximum value.
d) k(x) = -4 + x - x^2: The coefficient of x^2 is -1, the coefficient of x is 1, and the constant term is -4. Applying the formula, we get x = -1 / (2*(-1)) = 0.5. The vertex is (0.5, -4.25), and the function has a maximum value.
Based on these calculations, the correct answer is A) x^2 + 9, as it does not have a maximum value.
2. At what time will Myra's car be at its greatest height?
To determine the time when Myra's car is at its greatest height, we need to identify the maximum point of the height function h(t) = 10sin(6(t - 20°)) + 10. The maximum point occurs when the sine function reaches its maximum value of 1.
We have the equation 10sin(6(t - 20°)) + 10 = 11.
Subtracting 10 from both sides, we get 10sin(6(t - 20°)) = 1.
Dividing both sides by 10, we obtain sin(6(t - 20°)) = 0.1.
To find the angle whose sine is 0.1, we can use the arcsin function. arcsin(0.1) ≈ 5.74°.
Rearranging the equation, 6(t - 20°) = 5.74°.
Simplifying, t - 20° = 0.957°.
Adding 20° to both sides, we obtain t = 20.957°.
Given that t is in seconds, the answer is rounded to the nearest whole number, which is t = 21 seconds. Therefore, the correct answer is A) t = 20 s.
3. At which point on the graph of f(x) = -x^2 - 2x + 15 is the slope of the tangent 0?
To find the point on the graph of f(x) where the slope of the tangent is 0, we need to locate the x-coordinate where the derivative of the function is equal to 0.
Taking the derivative of f(x), we have f'(x) = -2x - 2.
Setting this derivative equal to 0, -2x - 2 = 0.
Solving for x, we get x = -1.
Plugging this value of x into the original function, f(-1) = -(-1)^2 - 2(-1) + 15 = 16.
Therefore, the point on the graph where the slope of the tangent is 0 is (-1, 16). Hence, the correct answer is option B) (-1,16).
To summarize:
1. The correct answer for the first question is A) x^2 + 9.
2. The correct answer for the second question is A) t = 20 s.
3. The correct answer for the third question is B) (-1,16).