Advanced Math

QUESTION 1 [2 marks] Easy

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

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$f(5)=5^2-4(5)+3=25-20+3=8$.

QUESTION 2 [2 marks] Easy

Solve: $x^2-9=0$.

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$x^2=9$, so $x=3$ or $x=-3$ (factoring: $(x-3)(x+3)=0$).

QUESTION 3 [3 marks] Medium

Solve: $x^2-5x+6=0$.

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Factor: $(x-2)(x-3)=0$, so $x=2$ or $x=3$.

QUESTION 4 [3 marks] Medium

The function $h(x)=2^x$. What is $h(3)-h(1)$?

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$h(3)=2^3=8$ and $h(1)=2^1=2$, so $h(3)-h(1)=8-2=6$.

QUESTION 5 [4 marks] Hard

The quadratic function $f(x)=ax^2+bx+c$ has zeros at $x=-2$ and $x=5$, and $f(0)=-20$. Find the value of $a$.

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Since the zeros are $-2$ and $5$, $f(x)=a(x+2)(x-5)$ for some constant $a$. Then $f(0)=a(0+2)(0-5)=a(2)(-5)=-10a$. Setting this equal to $-20$: $-10a=-20 \Rightarrow a=2$.

QUESTION 6 [5 marks] Hard

The expression $x^2-2x-8$ can be written as $(x+2)(x-c)$ for a constant $c$. What is the value of $c$?

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Expand $(x+2)(x-c)=x^2-cx+2x-2c=x^2+(2-c)x-2c$. Matching this to $x^2-2x-8$: the constant terms give $-2c=-8 \Rightarrow c=4$ (and checking the $x$-coefficient: $2-4=-2$ ?, confirming $c=4$).

QUESTION 7 [5 marks] Hard

A ball is thrown upward, and its height in meters after $t$ seconds is modeled by $h(t)=-5t^2+20t+2$. What is the maximum height the ball reaches, in meters?

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For a downward-opening parabola $h(t)=-5t^2+20t+2$, the maximum occurs at $t=-\dfrac{b}{2a}=-\dfrac{20}{2(-5)}=2$ seconds. The maximum height is $h(2)=-5(2)^2+20(2)+2=-20+40+2=22$ meters.

Nonlinear functions 7 questions