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Functions, 2ADV F1 2025 HSC 11

The graph of a quadratic function represented by the equation  \(h=t^2-8 t+12\)  is shown.
 

  1. Find the values of \(t\) and \(h\) at the turning point of the graph.   (2 marks)

  2. The graph shows  \(h=12\)  when  \(t=0\).
  3. What is the other value of \(t\) for which  \(h=12\)?   (1 mark)

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a.   \(\text{Turning point at} \ \ (4,-4)\)

b.   \(t=8\)

Show Worked Solution

a.    \(\text{Strategy 1 (no calculus)}\)

\(\text{Axis of quadratic occurs when}\ \ t= \dfrac{2+6}{2} = 4\)

\(\text{At} \ \ t=4:\)

\(h=4^2-8 \times 4+12=-4\)

\(\therefore \ \text{Turning point at} \ \ (4,-4)\)
 

\(\text{Strategy 2 (using calculus)}\)

\(h=t^2-8 t+12\)

\(h^{\prime}=2 t-8\)

\(\text{Find \(t\) when} \ \ h^{\prime}=0:\)

\(2 t-8=0 \ \Rightarrow \ t=4\)
 

b.    \(\text {When} \ \ h=12:\)

\(t^2-8 t+12\) \(=12\)
\(t(t-8)\) \(=0\)

 
\(\therefore \ \text{Other value:} \ \ t=8\)

Functions, 2ADV F1 EQ-Bank 1 MC

The graph of a quadratic function  \(f(x)=a x^2+b x+c\)  is drawn below.
 

Which of the following are true?

  1. \(a<0, c=0\)  and  \(b^2-4 a c=0\)
  2. \(a>0, c=0\)  and  \(b^2-4 a c=0\)
  3. \(a>0, c>0\)  and  \(b^2-4 a c>0\)
  4. \(a<0, c>0\)  and  \(b^2-4 a c=0\)
Show Answers Only

\(D\)

Show Worked Solution

\(\text{Quadratic touches } x \text{-axis once only} \ \ \Rightarrow b^2-4 a c=0\ \ \text{(eliminate C)}\)

\(\text{Quadratic is inverted} \Rightarrow a<0 \ \ \text{(eliminate B)}\)

\(\text{If} \ \ c=0, f(x)=a x^2+b x+0=x(a x+b) \Rightarrow \text{cuts twice (Eliminate A)}\)

\(\Rightarrow D\)

Functions, 2ADV F1 EQ-Bank 19

\(R\left(r, r^2\right), S\left(s, s^2\right)\) and \(T\left(t, t^2\right)\) are points on the parabola  \(y=x^2\).

Given \(RT\) is parallel to \(SO\), show  \(r+t=s\)   (2 marks)
 

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\(R\left(r, r^2\right), S\left(s, s^2\right), T\left(t, t^2\right)\)

\(m_{S O}=\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{s^2-0}{s-0}=s\)

\(m_{R T}=\dfrac{t^2-r^2}{t-r}=\dfrac{(t-r)(t+r)}{(t-r)}=t+r\)

\(\text{Given}\ R T \ \| \  SO \ \Rightarrow \ m_{SO}=m_{R T}\)

\(\therefore s=r+t\ \ …\ \text{as required} \)

Show Worked Solution

\(R\left(r, r^2\right), S\left(s, s^2\right), T\left(t, t^2\right)\)

\(m_{S O}=\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{s^2-0}{s-0}=s\)

\(m_{R T}=\dfrac{t^2-r^2}{t-r}=\dfrac{(t-r)(t+r)}{(t-r)}=t+r\)

\(\text{Given}\ R T \ \| \  SO \ \Rightarrow \ m_{SO}=m_{R T}\)

\(\therefore s=r+t\ \ …\ \text{as required} \)

Functions, 2ADV F1 EQ-Bank 17

The tangent to the parabola  \(y=x^2+2 x-4\)  is  \(y=px-5\)  where  \(p>0\).

Find the value of \(p\).   (2 marks)

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\(p=4\)

Show Worked Solution

\(\text{Intersection occurs when:}\)

\(x^2+2x-4\) \(=px-5\)  
\(x^2+(2-p)x+1\) \(=0\)  

 
\(\text{Tangent touches once}\ \Rightarrow\ \text{Discriminant}\ \Delta=0\)

\((2-p)^2-4 \times 1 \times 1\) \(=0\)  
\(4-4p+p^2-4\) \(=0\)  
\(p(p-4)\) \(=0\)  
\(p\) \(=4\ \ \ (p\gt 0)\)  
COMMENT: Key is to recognise this is a discriminant question, not a calculus application.

Functions, 2ADV F1 SM-Bank 25

Show that the parabola  \(2x^2-kx+k-2\)  has at least one real root.  (3 marks)

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\(2x^2-kx+k-2=0\)

\(\Delta=b^2-4ac=(-k)^2-4 \times 2(k-2) = k^{2}-8k+16\)

\(\text{Real roots:}\ \ \Delta \geqslant 0\)

\(k^2-8k+16\) \(\geqslant 0\)  
\((x-4)^2\) \(\geqslant 0\)  

 
\(\therefore\ \text{At least one root exists for all}\ k\)

Show Worked Solution

\(2x^2-kx+k-2=0\)

\(\Delta=b^2-4ac=(-k)^2-4 \times 2(k-2) = k^{2}-8k+16\)

\(\text{Real roots:}\ \ \Delta \geqslant 0\)

\(k^2-8k+16\) \(\geqslant 0\)  
\((x-4)^2\) \(\geqslant 0\)  

 
\(\therefore\ \text{At least one root exists for all}\ k\)

Functions, 2ADV F1 2023 HSC 10 MC

The graph  \(y = x^2\)  meets the line  \(y = k\)  (where \(k>0\)) at points \(P\) and \(Q\) as shown in the diagram. The length of the interval \(PQ\) is \(L\).
 

Let \(a\) be a positive number. The graph  \(y=\dfrac{x^2}{a^2}\)  meets the line  \(y=k\)  at points \(S\) and \(T\).

What is the length of \(ST\)?

  1. \(\dfrac{L}{a}\)
  2. \(\dfrac{L}{a^2}\)
  3. \(aL\)
  4. \(a^2L\)
Show Answers Only

\(C\)

Show Worked Solution

\(\text{Intersection of}\ \ y=x^2\ \ \text{and}\ \ y=k:\)

\(x^2=k\ \ \Rightarrow\ \ x=\pm \sqrt k\)

\(\therefore L=2\sqrt k\)

\(\text{Intersection of}\ \ y=\dfrac{x^2}{a^2}\ \ \text{and}\ \ y=k:\)

\(\dfrac{x^2}{a^2} \) \(=k\)  
\(x^2\) \(=a^2k\)  
\(x\) \(=\pm a\sqrt k\)  

\(\therefore ST=a \times 2\sqrt k = aL \)

\(\Rightarrow C\)

♦♦♦ Mean mark 24%.

Functions, 2ADV F1 SM-Bank 23

Find the values of `k` for which the expression  `x^2-3x + (4-2k)`  is always positive.  (3 marks)

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`k < 7/8`

Show Worked Solution

`x^2-3x + (4-2k) > 0`

`x^2-3x + (4-2k) = 0\ \ text(is a concave up parabola)`

`=>\ text{Always positive (no roots) if}\ \ Delta < 0`
 

`b^2-4ac < 0`

`(−3)^2-4 · 1 · (4-2k)` `< 0`
`9-16 + 8k` `< 0`
`8k` `< 7`
`k` `< 7/8`

Functions, 2ADV F1 2017 HSC 2 MC

Which expression is equal to  `3x^2-x-2`?

  1. `(3x-1) (x + 2)`
  2. `(3x + 1) (x-2)`
  3. `(3x-2) (x + 1)`
  4. `(3x + 2) (x-1)`
Show Answers Only

`D`

Show Worked Solution

`3x^2-x-2= (3x + 2) (x-1)`

`=>  D`

Functions, 2ADV F1 2016 HSC 11e

Find the points of intersection of  `y=-5-4x`  and  `y=3-2x-x^2.`  (3 marks)

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`(4, – 21) and (– 2, 3)`

Show Worked Solution

`y = 3 – 2x – x^2`

`text(Substitute)\ \ y = -5 – 4x\ \ text(into equation)`

`-5 – 4x` `= 3 – 2x – x^2`
`x^2 – 2x – 8` `= 0`
`(x – 4) (x + 2)` `= 0`

  
`:. x = 4 or -2`
 

`text(When)\ \ x = 4,\ \ y = -5 – 4(4) = -21`

`text(When)\ \ x = -2,\ \ y = -5 – 4 (-2) = 3`  
 

`:.\ text(Intersection at)\ \ (4, – 21) and (– 2, 3)`

Functions, 2ADV F1 2010 HSC 1a

Solve  `x^2 = 4x`.   (2 marks)

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 `x = 0\ text(or)\ 4`

Show Worked Solution
`x^2` `= 4x`
`x^2-4x` `= 0`
`x(x-4)` `= 0`

 

`:.\ x = 0\ text(or)\ 4`

Functions, 2ADV F1 2013 HSC 1 MC

What are the solutions of   `2x^2-5x-1 = 0`? 

  1. `x = (-5 +-sqrt17)/4` 
  2. `x = (5 +-sqrt17)/4`
  3. `x = (-5 +-sqrt33)/4`
  4. `x = (5 +-sqrt33)/4`
Show Answers Only

`D`

Show Worked Solution

`2x^2-5x-1 = 0`

`text(Using)\ x = (-b +- sqrt( b^2-4ac) )/(2a)`

`x` `= (5 +- sqrt{\ \ (-5)^2-4 xx 2 xx(-1) })/ (2 xx 2)`
  `= (5 +- sqrt(25 + 8) )/4`
  `= (5 +- sqrt(33) )/4`

 
`=>  D`