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A quadratic function, f_1, is defined on ℝ, the set of real numbers - Scottish Highers Maths - Question 15 - 2017

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Question 15

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A quadratic function, f_1, is defined on ℝ, the set of real numbers. Diagram 1 shows part of the graph with equation $y = f(x)$. The turning point is (2, 3). Diag... show full transcript

Worked Solution & Example Answer:A quadratic function, f_1, is defined on ℝ, the set of real numbers - Scottish Highers Maths - Question 15 - 2017

Step 1

Given that $h(x) = f(x + a) + b$. Write down the values of a and b.

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Answer

To find the values of a and b, we compare the turning points of the functions. The turning point of f(x)f(x) is at (2, 3) and for h(x)h(x) it is at (7, 6).

This gives us two equations based on the transformations:

  1. The x-coordinate of the turning point:

    2+a=7 a=72=52 + a = 7\ a = 7 - 2 = 5

  2. The y-coordinate of the turning point:

    3+b=6 b=63=33 + b = 6\ b = 6 - 3 = 3

Thus, the values are:

  • a=5a = 5
  • b=3b = 3

Step 2

It is known that $\int f'(x)dx = 4$. Determine the value of $\int h(x)dx$.

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Answer

Since we have determined that h(x)=f(x+5)+3h(x) = f(x + 5) + 3, we can find the integral of h(x)h(x) using a substitution method.

Let:

ext{If } u = x + 5 \ ext{Then } du = dx\ ext{And } \int h(x) dx = \int (f(u) + 3) du = \int f(u) du + \int 3 du. \end{align*}$$ Using the known integration: $$\int f'(x) dx = 4$$ The area under $h(x)$ will thus sum up to: $$\int h(x) dx = 4 + 3x$$ Representing this as: $$\int h(x) dx = 10$$ Therefore the value of $ esint{h(x)dx}$ is: 10.

Step 3

Given $f(1) = 6$, state the value of $h(8)$.

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Answer

To find h(8)h(8), we utilize the relationship:

h(x)=f(x+5)+3.h(x) = f(x + 5) + 3. Substituting 8 into the equation yields:

\text{Now we need to find } f(13).\$$ Since we don't have explicit values beyond what is derived, we conclude: $$h(8) = f(13) + 3$$ The exact numeric value is contingent on $f(13)$ but can be stated as the function value derived above.

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