SAT · SAT Maths
Algebra
Linear functions 7 questions
QUESTION 1 [2 marks]
Easy
A line has slope $3$ and passes through the point $(2,5)$. Write its equation in slope-intercept form.
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Using point-slope form: $y-5=3(x-2)$, so $y=3x-6+5=3x-1$.
QUESTION 2 [2 marks]
Easy
The function $f(x)=-2x+9$. Find $f(4)$.
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$f(4)=-2(4)+9=-8+9=1$.
QUESTION 3 [3 marks]
Medium
A line passes through $(1,4)$ and $(3,10)$. Find the equation of the line in slope-intercept form.
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Slope $=\dfrac{10-4}{3-1}=\dfrac{6}{2}=3$. Using point $(1,4)$: $y-4=3(x-1) \Rightarrow y=3x-3+4=3x+1$.
QUESTION 4 [3 marks]
Medium
The linear function $g$ satisfies $g(0)=7$ and $g(5)=22$. What is $g(10)$?
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Slope $=\dfrac{22-7}{5-0}=\dfrac{15}{5}=3$, and the $y$-intercept is $g(0)=7$, so $g(x)=3x+7$. Then $g(10)=3(10)+7=37$.
QUESTION 5 [4 marks]
Hard
Line $\ell$ has equation $4x-2y=10$. Line $m$ is perpendicular to $\ell$ and passes through $(3,-1)$. What is the $y$-intercept of line $m$?
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Rewrite $\ell$: $-2y=-4x+10 \Rightarrow y=2x-5$, so $\ell$ has slope $2$. A line perpendicular to $\ell$ has slope $-\dfrac{1}{2}$. Using point $(3,-1)$: $y-(-1)=-\dfrac{1}{2}(x-3) \Rightarrow y=-\dfrac{1}{2}x+\dfrac{3}{2}-1=-\dfrac{1}{2}x+\dfrac{1}{2}$. The $y$-intercept is $\dfrac{1}{2}$.
QUESTION 6 [5 marks]
Hard
The function $f$ is defined by $f(x)=ax+b$ for constants $a$ and $b$. If $f(3)=5$ and $f(7)=17$, what is the value of $f(0)$?
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The slope is $a=\dfrac{17-5}{7-3}=\dfrac{12}{4}=3$. Using $f(3)=5$: $3(3)+b=5 \Rightarrow 9+b=5 \Rightarrow b=-4$. So $f(x)=3x-4$, and $f(0)=b=-4$.
QUESTION 7 [5 marks]
Hard
Line $\ell$ passes through the points $(2,-1)$ and $(6,11)$. Line $m$ is parallel to $\ell$ and passes through the point $(0,5)$. What is the $x$-intercept of line $m$?
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Slope of $\ell$: $\dfrac{11-(-1)}{6-2}=\dfrac{12}{4}=3$. Since $m$ is parallel to $\ell$, it also has slope 3, and since $m$ passes through $(0,5)$, its equation is $y=3x+5$. The $x$-intercept is where $y=0$: $0=3x+5 \Rightarrow x=-\dfrac{5}{3}$.