Question Statement

Predict the functions from the following graphs:

(i) A line passing through points $\left(\frac{-1}{2},0\right)$ and $(0,1)$.

(ii) A parabola passing through points $(-1,1)$, $(0,-1)$, and $(1,1)$.

(iii) A parabola passing through points $(-2,-1)$, $(0,3)$, and $(2,-1)$.

(iv) A parabola passing through points $(-1,0)$ and $(0,1)$ with a given leading coefficient $a=1$.

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Background and Explanation

To determine a function from its graph, identify the shape of the curve to choose the correct general equation (linear: $y=mx+c$; quadratic: $y=a{x}^2+bx+c$). Substitute the coordinates of the given points into these equations to create a system of equations and solve for the unknown coefficients.

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Solution

Part (i)

The graph represents a straight line. We first find the slope ($m$) using the two given points $\left(\frac{-1}{2},0\right)$ and $(0,1)$.

$\begin{array}{rl}m& =\frac{y_2-y_1}{x_2-x_1}\\ & =\frac{1-0}{0-\left(\frac{-1}{2}\right)}\\ m& =\frac{1}{1/2}=2\end{array}$

Since the equation of the line is linear: \begin{align*} y &= mx + c \\ y &= 2x + c \end{align*}

Substitute the point $(0,1)$ into equation (1) to find the y-intercept ($c$): $\begin{array}{rl}1& =2(0)+c\\ c& =1\end{array}$

The final equation is: $\begin{array}{rl}y& =2x+1\\ 2x-y+1& =0\end{array}$

Part (ii)

The graph is a parabola. The general equation of a parabola is: $y=a{x}^2+bx+c$

Substitute the point $(0,-1)$ into equation (1): \begin{align*} -1 & = a(0)^2 + b(0) + c \Rightarrow c=-1 \\ y & =a x^{2}+b x-1 \end{align*}

Substitute the point $(-1,1)$ into equation (2): \begin{align*} 1 & =a(-1)^{2}+b(-1)-1 \\ 1+1 & =a-b \\ a & =b+2 \end{align*}

Substitute the point $(1,1)$ into equation (2): \begin{align*} 1 &= a(1)^{2}+b(1) \\ 1 &= a+b \end{align*}

Substitute the value of $a$ from equation (3) into equation (4): $\begin{array}{rl}1& =(b+2)+b\\ 2b& =1-2\\ 2b& =-1\\ b& =\frac{-1}{2}\end{array}$

Substitute $b$ back into equation (3): $\begin{array}{rl}a& =\frac{-1}{2}+2=\frac{-1+4}{2}\\ a& =\frac{3}{2}\end{array}$

Substitute $a$ and $b$ into equation (2): $\begin{array}{rl}y& =\frac{3}{2}{x}^2-\frac{1}{2}x-1\\ 2y& =3x^2-x-2\end{array}$

Part (iii)

The general equation of the parabola is: $y=a{x}^2+bx+c$

Substitute the point $(0,3)$ into equation (1): \begin{align*} 3 & = 0+0+c \\ c & = 3 \\ y & = a x^{2}+b x+3 \end{align*}

Substitute the point $(-2,-1)$ into equation (2): \begin{align*} -1 & =a(-2)^{2}+b(-2)+3 \\ -1-3 & =4 a-2 b \\ \frac{-4}{2} & =\frac{4a}{2}-\frac{2 b}{2} \\ -2 & =2 a-b \\ b & =2 a+2 \end{align*}

Substitute the point $(2,-1)$ into equation (2): $\begin{array}{rl}-1& =a(2{)}^2+b(2)+3\\ -1-3& =4a+2b\\ \frac{-4}{2}& =\frac{4a}{2}+\frac{2b}{2}\\ -2& =2a+b\end{array}$

Substitute $b=2a+2$ into the equation above: $\begin{array}{rl}-2& =2a+(2a+2)\\ -2-2& =4a\\ 4a& =-4\\ a& =-1\end{array}$

Substitute $a=-1$ into equation (3): $\begin{array}{rl}b& =2(-1)+2\\ b& =-2+2\\ b& =0\end{array}$ (Note: The raw data contains a slight calculation discrepancy in the final substitution for $b$, but the logic follows the substitution of $a$ into the derived expression for $b$.)

Using $a=-1,b=0,c=3$: $\begin{array}{rl}y& =(-1)x^2+(0)x+3\\ y& =-x^2+3\end{array}$

Part (iv)

Given $a=1$, the equation of the parabola is: \begin{align*} y &= ax^{2}+bx+c \\ y &= x^{2}+bx+c \end{align*}

Substitute the point $(-1,0)$ into equation (1): \begin{align*} 0 & =(-1)^{2}+b(-1)+c \\ 0 & =1-b+c \\ c & =b-1 \end{align*}

Substitute the point $(0,1)$ into equation (1): $\begin{array}{rl}1& =0+b(0)+c\\ c& =1\end{array}$

Substitute $c=1$ into equation (2) to find $b$: $\begin{array}{rl}1& =b-1\\ b& =1+1\\ b& =2\end{array}$

The final equation is: $y=x^2+2x+1$

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Key Formulas or Methods Used

• Slope Formula: $m=\frac{y_2-y_1}{x_2-x_1}$
• Linear Equation (Slope-Intercept Form): $y=mx+c$
• General Quadratic Equation: $y=a{x}^2+bx+c$
• Method of Substitution: Plugging known coordinates $(x,y)$ into the general equation to solve for constants $a,b,$ and $c$.

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Summary of Steps

1. Identify if the graph is a line or a parabola.
2. Write the general form of the equation ($y=mx+c$ or $y=a{x}^2+bx+c$).
3. Substitute the y-intercept point (where $x=0$) first, if available, to find the constant $c$ immediately.
4. Substitute the remaining points to create a system of linear equations for the remaining variables ($m$ or $a$ and $b$).
5. Solve the system using substitution or elimination.
6. Write the final function with the calculated coefficients.