The graph of $h(x) = ax + bx^2$ is drawn - NSC Mathematics - Question 11 - 2021 - Paper 1
Question 11
The graph of $h(x) = ax + bx^2$ is drawn.
The graph has turning points at the origin, O(0; 0) and B(4; 32).
A is an x-intercept of $h$.
Worked Solution & Example Answer:The graph of $h(x) = ax + bx^2$ is drawn - NSC Mathematics - Question 11 - 2021 - Paper 1
Step 1
Identify the Function and Its Components
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Answer
The function given is a quadratic in standard form, represented as h(x)=ax+bx2. The graph has turning points at (0, 0) and (4, 32), indicating values for h(0) and h(4).
Step 2
Determine the x-Intercept
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Answer
To find the x-intercept, we set the function h(x) equal to zero:
ax + bx^2 &= 0\
x(a + bx) &= 0.
\
\
This results in x=0 or a+bx=0, giving us the x-intercepts.
Step 3
Analyze the Turning Points
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Answer
The turning points provide critical information. The first turning point at the origin indicates that h(0)=0. The second turning point at (4, 32) suggests that:
h(4)=a(4)+b(42)=32.
Step 4
Setting Up the System of Equations
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Answer
From the previous analysis, we can form two equations:
h(0)=0⇒ax+bx2=0 leads us to the first intercept at x=0.
h(4)=32⇒4a+16b=32.
This system can now be solved to find values for a and b.
Step 5
Solve for Coefficients
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Answer
To solve for a and b, we rearrange:
4a+16b=32⇒a+4b=8.
We can substitute this back into our first equation to find the specific values. However, since the main focus is on the x-intercept, we note that the intercept occurs at x=0 for the quadratic, as derived earlier.
