Quadratic functions, equations and inequalities; discriminant – IB Maths exam questions with mark scheme

1StandardNon-calculator6 marks

Let $f(x)=2x^2-8x+5$.

(a) Write $f(x)$ in the form $a(x-h)^2+c$. [3]

(b) Write down the coordinates of the vertex of the graph of $f$. [1]

Factors out 2: $2(x^2-4x)+5$M1
$2(x-2)^2-8+5$A1
$2(x-2)^2-3$A1
Vertex $(2,-3)$A1
Discriminant of $2x^2-8x+5-k=0$: $64-8(5-k)>0$ (or minimum value $-3$)M1
$k>-3$A1
Answer: (a) $2(x-2)^2-3$ (b) $(2,-3)$ (c) $k>-3$
Examiner tip: Factor out the 2 from the $x^2$ and $x$ terms only; then the constant $2\times4$ is subtracted.

2HardNon-calculator6 marks

The graph of $y=x^2+(k-3)x+k$ touches the x-axis.

(a) Find the possible values of $k$. [4]

(b) For each value of $k$, find the coordinates of the point where the graph touches the x-axis. [2]

Touching means discriminant $=0$M1
$(k-3)^2-4k=0$ so $k^2-10k+9=0$A1
Factorises $(k-1)(k-9)=0$M1
$k=1$ or $k=9$A1
$k=1$: $x^2-2x+1=(x-1)^2$, point $(1,0)$A1
$k=9$: $x^2+6x+9=(x+3)^2$, point $(-3,0)$A1
Answer: (a) $k=1$ or $9$ (b) $(1,0)$ and $(-3,0)$
Examiner tip: Both values of $k$ are required; a quadratic in $k$ gives two answers.

3CoreNon-calculator6 marks

Let $f(x)=-(x+1)(x-5)$, for $x\in\mathbb{R}$.

(a) Write down the $x$-intercepts of the graph of $f$. [2]

(b) Write down the equation of the axis of symmetry of the graph of $f$. [1]

(d) Hence, or otherwise, find the range of $f$. [1]

$x=-1$ (accept $(-1,0)$)A1
$x=5$ (accept $(5,0)$)A1
$x=2$ (must be an equation; $2$ alone is A0)A1
Attempt to expand, e.g. $-(x^2-5x+x-5)$ or $-(x^2-4x-5)$M1
$f(x)=-x^2+4x+5$A1
$f(x)\le 9$ (accept $y\le9$, $]-\infty,9]$; do not accept $x\le9$ or $f(x)<9$)A1
Answer: (a) $x=-1,\ x=5$ (b) $x=2$ (c) $-x^2+4x+5$ (d) $f(x)\le 9$
Examiner tip: The range is about output values: write $f(x)\le9$, not $x\le9$, and use $\le$ because the vertex value is attained.

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4CoreNon-calculator5 marks

(a) Solve the inequality $3x^2+5x-2>0$. [4]

(b) Hence write down the set of values of $x$ for which $3x^2+5x-2\le 0$. [1]

Valid attempt to find the critical values, e.g. $(3x-1)(x+2)$ or quadratic formulaM1
Critical values $x=\tfrac13$ and $x=-2$A1
Valid method to choose regions, e.g. sketch of a positive parabola, sign diagram or test valuesM1
$x<-2$ or $x>\tfrac13$ (accept $x\in\,]-\infty,-2[\,\cup\,]\tfrac13,\infty[$; A0 for $\tfrac13A1
$-2\le x\le\tfrac13$ (follow through from their critical values; must include equality)A1
Answer: (a) $x<-2$ or $x>\tfrac13$ (b) $-2\le x\le \tfrac13$
Examiner tip: Finding the roots is only half the job: a quick sketch of the U-shaped graph shows the curve is above the axis outside the roots.

5StandardNon-calculator8 marks

The diagram shows part of the graph of a quadratic function $y=f(x)$. The graph crosses the $x$-axis at $(-1,0)$ and $(3,0)$, and crosses the $y$-axis at $(0,-6)$.

(a) Find $f(x)$, giving your answer in the form $a(x-p)(x-q)$. [3]

(b) Find the coordinates of the vertex of the graph. [2]

$p=-1,\ q=3$ seen, e.g. $a(x+1)(x-3)$A1
Substitutes $(0,-6)$: $a(1)(-3)=-6$M1
$f(x)=2(x+1)(x-3)$A1
$x=1$ (from symmetry or completing the square)A1
Vertex $(1,-8)$A1
Forms a quadratic, e.g. $2x^2-4x-6>10$ leading to $x^2-2x-8>0$M1
Critical values $x=4,\ x=-2$A1
$x<-2$ or $x>4$A1
Answer: (a) $f(x)=2(x+1)(x-3)$ (b) $(1,-8)$ (c) $x<-2$ or $x>4$
Examiner tip: The intercepts fix $p$ and $q$ but not the stretch factor: always use the third point to find $a$.

6StandardCalculator7 marks

A ball is thrown upwards from the edge of a platform. Its height above the ground, $h$ metres, $t$ seconds after it is thrown, is modelled by

$$h(t)=-4.9t^2+14t+1.5,\quad t\ge0.$$

(a) Find the time at which the ball reaches its maximum height. [2]

(b) Find the maximum height of the ball. [1]

(d) Find the length of time for which the ball is more than $8$ metres above the ground. [2]

Uses $t=-\dfrac{b}{2a}=\dfrac{-14}{2(-4.9)}$ or GDC maximumM1
$t=1.43$ s ($1.42857\ldots$, accept $\tfrac{10}{7}$)A1
$11.5$ mA1
Attempt to solve $h(t)=0$, e.g. GDC or quadratic formulaM1
$t=2.96$ s ($2.96054\ldots$; reject the negative root; A0 if both roots given as final answer)A1
Attempt to solve $h(t)=8$: roots $0.583\ldots$ and $2.27\ldots$M1
$1.69$ s ($1.69031\ldots$)A1
Answer: (a) $1.43$ s (b) $11.5$ m (c) $2.96$ s (d) $1.69$ s
Examiner tip: Give the context-sensible root only in (c), and in (d) subtract the two times rather than quoting them.

7StandardNon-calculator6 marks

The line $y=mx-1$ is a tangent to the curve $y=x^2+3$.

(a) Show that the $x$-coordinate of the point of contact satisfies $x^2-mx+4=0$. [1]

(b) Find the possible values of $m$. [3]

Equates and rearranges: $x^2+3=mx-1\Rightarrow x^2-mx+4=0$ (AG)M1
Recognises tangency means equal roots: $\Delta=0$M1
$m^2-16=0$ (accept $(-m)^2-4(1)(4)=0$)A1
$m=\pm4$A1
Substitutes $m=4$: $x^2-4x+4=0$, so $(x-2)^2=0$M1
$(2,7)$A1
Answer: (a) shown (b) $m=4$ or $m=-4$ (c) $(2,7)$
Examiner tip: Tangent means exactly one intersection, so set the discriminant equal to zero, not greater than zero.

8HardNon-calculator6 marks

Consider the equation $x^2+(k+1)x+k-1=0$, where $k\in\mathbb{R}$.

(a) Show that the equation has two distinct real roots for all values of $k$. [4]

(b) One root of the equation is $0$. Find the value of $k$ and the other root. [2]

Attempt to find the discriminant: $(k+1)^2-4(1)(k-1)$M1
$k^2-2k+5$A1
Attempt to complete the square: $(k-1)^2+4$M1
States $(k-1)^2\ge0$ so $\Delta\ge4>0$ for all $k$, hence two distinct real roots (AG)R1
$k-1=0\Rightarrow k=1$A1
Other root $x=-2$ (from $x^2+2x=0$)A1
Answer: (a) $\Delta=(k-1)^2+4>0$ (b) $k=1$, other root $-2$
Examiner tip: For a 'show that' the reasoning mark needs the explicit statement that a square is never negative, so the discriminant is always positive.

9HardNon-calculator7 marks

Let $g(x)=(k-2)x^2+4x+k+1$, where $k\in\mathbb{R}$ and $k\ne2$.

Find the set of values of $k$ for which $g(x)<0$ for all $x\in\mathbb{R}$. [7]

Recognises the graph must open downwards: $k-2<0$M1
$k<2$A1
Recognises there must be no real roots: $\Delta<0$M1
$16-4(k-2)(k+1)<0$A1
Simplifies to $k^2-k-6>0$ (or $-4k^2+4k+24<0$)A1
$k<-2$ or $k>3$A1
Combines both conditions: $k<-2$A1
Answer: $k<-2$
Examiner tip: Two conditions are needed: no real roots AND a negative leading coefficient; forgetting the second leaves $k>3$ wrongly in the answer.

10HardNon-calculatorHL only7 marks

The equation $2x^2-6x+1=0$ has roots $\alpha$ and $\beta$.

(a) Write down the value of $\alpha+\beta$ and the value of $\alpha\beta$. [2]

(b) Find the value of $\alpha^2+\beta^2$. [2]

$\alpha+\beta=3$A1
$\alpha\beta=\tfrac12$A1
Uses $\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta=9-1$M1
$8$A1
Product of new roots $\alpha^2\beta^2=(\alpha\beta)^2=\tfrac14$A1
Forms $x^2-(\text{sum})x+(\text{product})=0$, e.g. $x^2-8x+\tfrac14=0$M1
$4x^2-32x+1=0$ (accept any integer multiple)A1
Answer: (a) $\alpha+\beta=3,\ \alpha\beta=\tfrac12$ (b) $8$ (c) $4x^2-32x+1=0$
Examiner tip: Sum of roots is $-b/a$, not $b/a$: check the sign before using it in every later part.

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