Graphs of Functions: past paper questions & worked solutions

The graph has x-intercepts $(-4,0)$ and $\left(\dfrac{k}{2},0\right)$. The root at $x=-4$ is a repeated root, so the graph touches the x-axis there.

Since $0<k<4$, the positive intercept $x=\dfrac{k}{2}$ lies between 0 and 2. The y-intercept is $(0,-16k)$.

The graph is positive on both sides of the y-axis, with asymptotes $x=0$ and $y=0$. It has no intercepts with the coordinate axes.

The equation $(2x-k)(x+4)^2=\dfrac{k}{x^2}$ represents intersections between the two graphs.

From the sketches, the positive branch of $\dfrac{k}{x^2}$ meets the increasing positive branch of $(2x-k)(x+4)^2$ once only. Therefore there is one root.

$y=x(4-x^2)$, so $y=0$ when $x=0$ or $x=\pm2$. The curve crosses the axes at $(-2,0)$, $(0,0)$ and $(2,0)$.

$y=4x-x^3$ has negative leading cubic term, so the left tail rises and the right tail falls.

$x(4-x^2)=\dfrac{A}{x}$.

$x^2(4-x^2)=A$, so $4x^2-x^4=A$ and therefore $x^4-4x^2+A=0$.

Let $u=x^2$. Then $u^2-4u+A=0$.

For 4 distinct x-values, the quadratic in $u$ must have two distinct positive roots.

The discriminant gives $16-4A>0$, so $A<4$. The product of the two roots is $A$, so $A>0$. Therefore $0<A<4$.

For $y=\dfrac{1}{x+2}$, the denominator is zero when $x=-2$, so the vertical asymptote is $x=-2$.

$x^3+4x^2+4x=x(x^2+4x+4)=x(x+2)^2$.

$x=0$ is a single root, so the curve crosses the x-axis at $(0,0)$. $x=-2$ is a repeated root, so the curve touches the x-axis at $(-2,0)$.

$(x+2)(x^3+4x^2+4x)=1$ is equivalent to $x^3+4x^2+4x=\dfrac{1}{x+2}$.

The number of real solutions is the number of intersections of the cubic $y=x(x+2)^2$ and the reciprocal curve $y=\dfrac{1}{x+2}$. The sketches show 2 intersections.

The line $x+y=6$ has equation $y=6-x$. Intersections satisfy $6-x=6x-2x^2+1$.

Rearranging gives $2x^2-7x+5=0$, so $(2x-5)(x-1)=0$. Therefore $x=1$ or $x=\dfrac52$.

Using $y=6-x$, the points are $P=(1,5)$ and $Q=\left(\dfrac52,\dfrac72\right)$.

Region $R$ is above the x-axis, below the line, and above the curve.

Thus $y\ge0$, $y\le6-x$, and $y\ge6x-2x^2+1$.

$y=\dfrac2x-k$ has horizontal asymptote $y=-k$.

Setting $y=0$ gives $\dfrac2x=k$, so $x=\dfrac2k$. There is no y-intercept because $x=0$ is not in the domain.

$\dfrac2x-k=-kx-6$.

Multiplying by $x$ gives $2-kx=-kx^2-6x$, so $kx^2+(6-k)x+2=0$.

For two distinct points, the discriminant must be positive.

$(6-k)^2-8k>0$, so $k^2-20k+36>0$. Since $k^2-20k+36=(k-2)(k-18)$, $k<2$ or $k>18$.

As $k>0$, the range is $0<k<2$ or $k>18$.

$P_A(0)=53-0.4(0-8)^2=53-25.6=27.4$. Also $P_B(0)=-1.6(0)+44.2=44.2$.

The difference is $44.2-27.4=16.8$, so the share values differ by $16.8$ million pounds.

$P_A=53-0.4(t-8)^2$ has maximum value $53$, reached when $t=8$, which is within the 15-year period.

$53-0.4(t-8)^2>-1.6t+44.2$.

Expanding gives $-0.4t^2+8t-16.8>0$. Multiplying by $-2.5$ reverses the inequality: $t^2-20t+42<0$.

$t=\dfrac{20\pm\sqrt{400-168}}{2}=10\pm\sqrt{58}$. Therefore company A is greater between the two roots.

Since the model is monitored over $0\le t\le15$, the required times are $10-\sqrt{58}<t\le15$.

The model should not be used at $t=20$ because it was only built from a 15-year monitoring period; $20$ years is outside that range.

$f(x)=(4x-3)(x-5)^2$. The curve crosses the x-axis at $x=\frac34$ and touches the x-axis at $x=5$. The y-intercept is $f(0)=(-3)(25)=-75$.

For $f\left(\frac14x\right)=0$, either $4\left(\frac{x}{4}\right)-3=0$ or $\frac{x}{4}-5=0$. Therefore $x=3$ or $x=20$.

The curve $y=f(x)+p$ passes through the origin, so $0=f(0)+p=-75+p$. Hence $p=75$.

$g(x)=f(x+1)=(4(x+1)-3)((x+1)-5)^2=(4x+1)(x-4)^2$.

At $x=0$, $g(0)=(1)(16)=16$. So the y-intercept is $16$.

For $y=\frac4x+k$, the horizontal asymptote is $y=k$. Setting $y=0$ gives $0=\frac4x+k$, so the x-intercept is $x=-\frac4k$.

A tangent touches once, so solve $10-2x=\frac4x+k$. Multiplying by $x$ gives $2x^2+(k-10)x+4=0$.

For one repeated intersection, discriminant $=0$: $(k-10)^2-4(2)(4)=0$. Hence $(k-10)^2=32$.

$k-10=\pm4\sqrt2$, so $k=10\pm4\sqrt2$. Both are positive, so both satisfy the condition in the question.

$y=\dfrac{4k}{x-2k}$ has no x-intercept because the numerator is non-zero. At $x=0$, $y=\dfrac{4k}{-2k}=-2$, so the y-intercept is $(0,-2)$.

The denominator is zero when $x=2k$, so the vertical asymptote is $x=2k$.

The curve has one branch in quadrant I to the right of $x=2k$, and one branch passing through $(0,-2)$ and approaching the x-axis from below as $x\to-\infty$.

$\dfrac{4k}{x-2k}=6-2x$.

$4k=(6-2x)(x-2k)=6x-12k-2x^2+4kx$, so $2x^2-(4k+6)x+16k=0$.

For at least one real intersection, the discriminant must satisfy $b^2-4ac\ge0$.

$(4k+6)^2-4(2)(16k)\ge0$, so $16k^2-80k+36\ge0$.

$4k^2-20k+9\ge0$, so $(2k-1)(2k-9)\ge0$.

Therefore $k\le\dfrac12$ or $k\ge\dfrac92$. Since $k$ is positive, $0<k\le\dfrac12$ or $k\ge\dfrac92$.

$y=\dfrac1{2-x}$ has vertical asymptote $x=2$ and horizontal asymptote $y=0$.

At $x=0$, $y=\dfrac12$, so the y-intercept is $(0,\dfrac12)$.

If $y=kx-4$ cuts $C$, then $kx-4=\dfrac1{2-x}$.

$(kx-4)(2-x)=1$, so $kx^2-(2k+4)x+9=0$.

For at least one intersection, the discriminant is non-negative.

$(2k+4)^2-4(k)(9)\ge0$, so $4(k^2-5k+4)\ge0$, hence $k^2-5k+4\ge0$.

$k^2-5k+4=(k-1)(k-4)$, so $k\le1$ or $k\ge4$.

For $y=\dfrac{4}{x-k}$, the vertical asymptote is where the denominator is zero, so $x=k$.

At $x=0$, $y=-\dfrac4k$, so the y-intercept is $(0,-\dfrac4k)$. The graph has no x-intercept because the numerator is never zero.

If the line $y=9-x$ meets the curve, then $9-x=\dfrac4{x-k}$.

$(9-x)(x-k)=4$, which simplifies to $x^2-(9+k)x+9k+4=0$.

For the line not to cross or touch the curve, the discriminant must be negative.

$(9+k)^2-4(9k+4)<0$, so $k^2-18k+65<0$. Since $k^2-18k+65=(k-5)(k-13)$, the range is $5<k<13$.

For x-intercepts, set $\dfrac1{x^2}-9=0$. Then $x^2=\dfrac19$, so $x=\pm\dfrac13$. There is no y-intercept because $x=0$ is not in the domain.

The vertical asymptote is $x=0$. As $|x|$ becomes large, $\dfrac1{x^2}\to0$, so the horizontal asymptote is $y=-9$.

For intersections with $y=kx^2$, solve $\dfrac1{x^2}-9=kx^2$.

Let $u=x^2$, where $u>0$. Multiplying by $x^2$ gives $1-9u=ku^2$, so $ku^2+9u-1=0$.

Four distinct x-values require two distinct positive values of $u$, since each positive $u$ gives $x=\pm\sqrt{u}$.

For two positive roots, the product $-\dfrac1k$ and sum $-\dfrac9k$ must both be positive, so $k<0$. Also the discriminant must be positive: $81+4k>0$, so $k>-\dfrac{81}{4}$. Therefore $-\dfrac{81}{4}<k<0$.

$f(x)=(3x-2)^2(x-4)$. The squared factor is never negative, and it is zero at $x=\dfrac23$. Therefore the sign of $f(x)$ away from the repeated root is controlled by $x-4$.

For $f(x)>0$, we need $x-4>0$, so $x>4$.

$(3x-2)^2=9x^2-12x+4$. Therefore $f(x)=(9x^2-12x+4)(x-4)=9x^3-48x^2+52x-16$.

The y-intercept of $C$ is $f(0)=-16$, so the horizontal line $l$ is $y=-16$.

Solve $f(x)=-16$: $9x^3-48x^2+52x-16=-16$, so $x(9x^2-48x+52)=0$. The root $x=0$ is the y-intercept, and the two other roots give $P$ and $Q$.

$9x^2-48x+52=0$. Hence $x=\dfrac{48\pm\sqrt{48^2-4(9)(52)}}{18}=\dfrac{48\pm12\sqrt3}{18}=\dfrac{8\pm2\sqrt3}{3}$.

Because $P$ and $Q$ lie on the same horizontal line, $PQ$ is the difference in their x-coordinates: $PQ=\dfrac{8+2\sqrt3}{3}-\dfrac{8-2\sqrt3}{3}=\dfrac{4\sqrt3}{3}$.

Since the cubic touches the x-axis at the origin and cuts it at $x=4$, write $f(x)=\lambda x^2(x-4)$.

$120=\lambda(10)^2(10-4)=600\lambda$, so $\lambda=0.2$.

$f(x)=0.2x^2(x-4)$.

$C_2:y=1.2x(8-x)$ is a downward-opening parabola through $x=0$ and $x=8$, with its maximum between them.

$1.2x(8-x)=0.2x^2(x-4)$.

Divide by $0.2$ to get $6x(8-x)=x^2(x-4)$.

$48x-6x^2=x^3-4x^2$, so $x^3+2x^2-48x=0$ and $x(x^2+2x-48)=0$.

$x=0$, or $x^2+2x-48=0=(x+8)(x-6)$, so $x=-8$ or $x=6$.

Using $y=1.2x(8-x)$ gives $(-8,-153.6)$, $(0,0)$ and $(6,14.4)$.

$f(x)=2(x+1)(x-3)^2$. The curve cuts the x-axis at $(-1,0)$ and meets the x-axis at $(3,0)$ because $x=3$ is a repeated root.

$f(0)=2(1)(9)=18$, so the y-intercept is $(0,18)$.

The leading term is positive, so the cubic goes down to the left and up to the right. It crosses at $x=-1$ and touches at $x=3$.

$(x+1)(x-3)^2=(x+1)(x^2-6x+9)=x^3-5x^2+3x+9$, so $f(x)=2x^3-10x^2+6x+18$.

$f'(x)=6x^2-20x+6$.

$f'\left(\dfrac13\right)=6\left(\dfrac19\right)-20\left(\dfrac13\right)+6=0$, so the tangent is horizontal.

$f\left(\dfrac13\right)=2\left(\dfrac43\right)\left(-\dfrac83\right)^2=\dfrac{512}{27}$.

A horizontal tangent through this point has equation $y=\dfrac{512}{27}$.

Because $k>0$, $y=-\dfrac{k}{x}$ has one branch in quadrant II and one branch in quadrant IV, with asymptotes $x=0$ and $y=0$.

$y=-\dfrac{k}{x}+k$ is $y=-\dfrac{k}{x}$ translated up by $k$ units. The vertical asymptote is still $x=0$, and the horizontal asymptote is $y=k$.

Set $y=0$: $0=-\dfrac{k}{x}+k$. Since $k>0$, $\dfrac{k}{x}=k$, so $x=1$. The x-intercept is $(1,0)$.

For intersections with $y=3x+4$, solve $-\dfrac{k}{x}+k=3x+4$. Multiplying by $x$ gives $-k+kx=3x^2+4x$.

This rearranges to $3x^2+(4-k)x+k=0$. For no touch or intersection, the discriminant must be negative: $(4-k)^2-12k<0$.

$k^2-20k+16<0$. The roots are $k=10\pm2\sqrt{21}$, so the required range is $10-2\sqrt{21}<k<10+2\sqrt{21}$.

$y=x^2-5x+13=\left(x-\frac52\right)^2+\frac{27}{4}$. Therefore $M=\left(\frac52,\frac{27}{4}\right)$.

The gradient of $OM$ is $\dfrac{27/4}{5/2}=\dfrac{27}{10}$, so $l$ has equation $y=\frac{27}{10}x$.

Solve $x^2-5x+13=\frac{27}{10}x$. This gives $x^2-\frac{77}{10}x+13=0$. One root is $x=\frac52$, so the other is $x=\frac{26}{5}$. Then $y=\frac{27}{10}\cdot\frac{26}{5}=\frac{351}{25}$.

The region is between the y-axis, the line, and the curve up to $M$: $0\le x\le\frac52$ and $\frac{27}{10}x\le y\le x^2-5x+13$.

$f(x)=(2x+5)(x-3)^2$. Since $(x-3)^2\ge0$, the sign is controlled by $2x+5$, with an extra zero at $x=3$. Hence $f(x)\le0$ for $x\le-\frac52$ or $x=3$.

$(x-3)^2=x^2-6x+9$, so $f(x)=(2x+5)(x^2-6x+9)=2x^3-7x^2-12x+45$.

At the y-axis, $x=0$, so $P=(0,45)$. Also $f'(x)=6x^2-14x-12$, so the gradient at $P$ is $f'(0)=-12$.

A translation 2 units in the positive x direction means $g(x)=f(x-2)$. Therefore $g(x)=(2(x-2)+5)((x-2)-3)^2=(2x+1)(x-5)^2$.

Set $x=0$: $g(0)=(1)(25)=25$, so the y-intercept is $25$.

The curve $$y=\frac{3k}{x-2k}$$ is a rectangular hyperbola with vertical asymptote $x=2k$ and horizontal asymptote $y=0$. Since $k>0$, the branch for $x>2k$ lies above the x-axis. The other branch lies below the x-axis and passes through the y-axis.

At $x=0$, $$y=\frac{3k}{-2k}=-\frac32.$$ So the curve crosses the y-axis at $$\left(0,-\frac32\right).$$ There is no x-intercept because the numerator $3k$ is non-zero.

The line is $y=12-x$. At an intersection, $$\frac{3k}{x-2k}=12-x.$$ Cross-multiplying gives $$3k=(12-x)(x-2k)=12x-x^2-24k+2kx.$$ Hence $$x^2-(12+2k)x+27k=0.$$

A tangent gives a repeated solution for $x$, so the discriminant is zero: $$(12+2k)^2-4(1)(27k)=0.$$ Expanding and simplifying, $$144+48k+4k^2-108k=0,$$ $$4k^2-60k+144=0,$$ and therefore $$k^2-15k+36=0,$$ as required.

Factor the equation from part (b): $$k^2-15k+36=(k-3)(k-12)=0.$$ Thus $k=3$ or $k=12$.

The repeated root of $$x^2-(12+2k)x+27k=0$$ is $$x=\frac{12+2k}{2}=6+k.$$ Then $y=12-x=6-k$. For $k=3$, the point is $(9,3)$, which is above the x-axis. For $k=12$, the point is $(18,-6)$, which is below the x-axis. Therefore $$P=(18,-6).$$