Home / IBDP Maths AA: Topic: SL 2.6: The quadratic function: IB style Questions HL Paper 2

IBDP Maths AA: Topic: SL 2.6: The quadratic function: IB style Questions HL Paper 2

Question

Let \(f\) be a quadratic function. Part of the graph of \(f\) is shown below.

The vertex is at P(\(4\), \(2\)) and the y-intercept is at Q(\(0\), \(6\)) .

a.Write down the equation of the axis of symmetry.[1]

b.The function f can be written in the form \(f(x) = a{(x – h)^2} + k\) .

Write down the value of h and of k .[2]

c.The function f can be written in the form \(f(x) = a{(x – h)^2} + k\) .

Find a .[3]

 
▶️Answer/Explanation

Markscheme

\(x = 4\) (must be an equation)     A1     N1

[1 mark]

a.

\(h = 4\) , \(k = 2\)     A1A1     N2

[2 marks]

b.

attempt to substitute coordinates of any point on the graph into f     (M1)

e.g. \(f(0) = 6\) , \(6 = a{(0 – 4)^2} + 2\) , \(f(4) = 2\)

correct equation (do not accept an equation that results from \(f(4) = 2\) )     (A1)

e.g. \(6 = a{( – 4)^2} + 2\) , \(6 = 16a + 2\)

\(a = \frac{4}{{16}}\left( { = \frac{1}{4}} \right)\)     A1     N2

[3 marks]

c.

Question

Let \(f(x) = \sin x + \frac{1}{2}{x^2} – 2x\) , for \(0 \le x \le \pi \) .

Let \(g\) be a quadratic function such that \(g(0) = 5\) . The line \(x = 2\) is the axis of symmetry of the graph of \(g\) .

The function \(g\) can be expressed in the form \(g(x) = a{(x – h)^2} + 3\) .

a.Find \(f'(x)\) .[3]

b.Find \(g(4)\) .[3]
c.(i)     Write down the value of \(h\) .

(ii)     Find the value of \(a\) .[4]

d.Find the value of \(x\) for which the tangent to the graph of \(f\) is parallel to the tangent to the graph of \(g\) .[6]
 
▶️Answer/Explanation

Markscheme

\(f'(x) = \cos x + x – 2\)     A1A1A1     N3

Note: Award A1 for each term.

[3 marks]

a.

recognizing \(g(0) = 5\) gives the point (\(0\), \(5\))     (R1)

recognize symmetry     (M1)

eg vertex, sketch

\(g(4) = 5\)     A1     N3

[3 marks]

b.

(i)     \(h = 2\)     A1 N1

(ii)     substituting into \(g(x) = a{(x – 2)^2} + 3\) (not the vertex)     (M1)

eg   \(5 = a{(0 – 2)^2} + 3\) , \(5 = a{(4 – 2)^2} + 3\)

working towards solution     (A1)

eg   \(5 = 4a + 3\) , \(4a = 2\)

\(a = \frac{1}{2}\)     A1     N2

[4 marks]

c.

\(g(x) = \frac{1}{2}{(x – 2)^2} + 3 = \frac{1}{2}{x^2} – 2x + 5\)

correct derivative of \(g\)     A1A1

eg   \(2 \times \frac{1}{2}(x – 2)\) , \(x – 2\)

evidence of equating both derivatives     (M1)

eg   \(f’ = g’\)

correct equation     (A1)

eg   \(\cos x + x – 2 = x – 2\)

working towards a solution     (A1)

eg   \(\cos x = 0\) , combining like terms

\(x = \frac{\pi }{2}\)    A1     N0

Note: Do not award final A1 if additional values are given.

[6 marks]

Question

Let \(f(x) = a{(x – h)^2} + k\). The vertex of the graph of \(f\) is at \((2, 3)\) and the graph passes through \((1, 7)\).

a.Write down the value of \(h\) and of \(k\).[2]

 

b.Find the value of \(a\).[3]

 
▶️Answer/Explanation

Markscheme

\(h = 2,{\text{ }}k = 3\)     A1A1     N2

[2 marks]

a.

attempt to substitute \((1,7)\) in any order into their \(f(x)\)     (M1)

eg     \(7 = a{(1 – 2)^2} + 3{\text{, }}7 = a{(1 – 3)^2} + 2{\text{, }}1 = a{(7 – 2)^2} + 3\)

correct equation     (A1)

eg     \(7 = a + 3\)

a = 4     A1     N2

[3 marks]

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