Question Statement

Draw the graph of the following functions using factors:

(i) $f(x)=x^2-2x+1$ (ii) $f(x)=x^2-7x+12$ (iii) $f(x)=x^2-2x$ (iv) $f(x)=-2x^2+x+3$ (v) $f(x)=4x^2-4x$ (vi) $f(x)=6-x^2-x$

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Background and Explanation

To graph a quadratic function $f(x)=a{x}^2+bx+c$ using factors, we first find the $x$-intercepts by setting $f(x)=0$. The vertex's $x$-coordinate is the midpoint of these intercepts, and the sign of the coefficient $a$ determines if the parabola opens upwards ($a>0$) or downwards ($a<0$).

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Solution

Part (i)

Given the function: $f(x)=x^2-2x+1$

Here $a=1,b=-2,c=1$. To find the factors and $x$-intercepts, we set $f(x)=0$: $\begin{array}{rl}{x}^2-2x+1& =0\\ (x-1{)}^2& =0\\ x-1& =0\\ x& =1\end{array}$

The graph touches the $x$-axis at the point $(1,0)$. To find the $y$-intercept, we set $x=0$: $f(0)=0-0+1=1$ Therefore, the $y$-intercept is $(0,1)$.

Vertex: The $x$-coordinate of the vertex is the average of the roots: $x\text{-coordinate}=\frac{1+1}{2}=1$ Substituting $x=1$ into the function to find the $y$-coordinate: $y=1-2+1=0$ The vertex is $(1,0)$. Since $a>0$, the parabola opens upwards and is symmetric about the line $x=1$.

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Part (ii)

Given the function: $f(x)=x^2-7x+12$

Here $a=1,b=-7,c=12$. To find the factors, we set $f(x)=0$: $\begin{array}{rl}{x}^2-7x+12& =0\\ x^2-3x-4x+12& =0\\ x(x-3)-4(x-3)& =0\\ (x-3)(x-4)& =0\end{array}$ So, $x-3=0⟹x=3$ or $x-4=0⟹x=4$. The graph intersects the $x$-axis at $(3,0)$ and $(4,0)$.

To find the $y$-intercept, set $x=0$: $f(0)=12$ The $y$-intercept is $(0,12)$.

Vertex: $x\text{-coordinate}=\frac{3+4}{2}=3.5$ Substituting $x=3.5$ into the function: $\begin{array}{rl}y\text{-coordinate}& =(3.5{)}^2-7(3.5)+12\\ & =12.25-24.5+12\\ & =-0.25\end{array}$ The vertex is $(3.5,-0.25)$. Since $a>0$, the parabola opens upwards and is symmetric about $x=3.5$.

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Part (iii)

Given the function: $f(x)=x^2-2x$

Here $a=1,b=-2,c=0$. To find the factors, we set $f(x)=0$: $\begin{array}{rl}{x}^2-2x& =0\\ x(x-2)& =0\end{array}$ So, $x=0$ or $x-2=0⟹x=2$. The graph intersects the $x$-axis at $(0,0)$ and $(2,0)$.

To find the $y$-intercept, set $x=0$: $f(0)=0-0=0$ The $y$-intercept is $(0,0)$.

Vertex: $x\text{-coordinate}=\frac{0+2}{2}=1$ Substituting $x=1$ into the function: $f(1)=1^2-2(1)=-1$ The vertex is $(1,-1)$. Since $a>0$, the parabola opens upwards and is symmetric about $x=1$.

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Part (iv)

Given the function: $f(x)=-2x^2+x+3$

Here $a=-2,b=1,c=3$. To find the factors, we set $f(x)=0$: $\begin{array}{rl}-2x^2+x+3& =0\\ 2x^2-x-3& =0\\ 2x^2-3x+2x-3& =0\\ x(2x-3)+1(2x-3)& =0\\ (x+1)(2x-3)& =0\end{array}$ So, $x+1=0⟹x=-1$ or $2x-3=0⟹x=\frac{3}{2}$. The graph intersects the $x$-axis at $(-1,0)$ and $(\frac{3}{2},0)$.

To find the $y$-intercept, set $x=0$: $f(0)=3$ The $y$-intercept is $(0,3)$.

Vertex: $x\text{-coordinate}=\frac{-1+\frac{3}{2}}{2}=\frac{1/2}{2}=\frac{1}{4}=0.25$ Substituting $x=\frac{1}{4}$ into the function: $y\text{-coordinate}=-2{\left(\frac{1}{4}\right)}^2+\frac{1}{4}+3=3.125$ The vertex is $(0.25,3.125)$. Since $a<0$, the parabola opens downwards.

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Part (v)

Given the function: $f(x)=4x^2-4x$

Here $a=4,b=-4,c=0$. To find the factors, we set $f(x)=0$: $\begin{array}{rl}4x^2-4x& =0\\ 4x(x-4)& =0\\ x=0& \text{ or }x=4\end{array}$ (Note: Based on raw data steps, the root $x=4$ is used for the vertex calculation). The graph intersects the $x$-axis at $(0,0)$ and $(4,0)$.

To find the $y$-intercept, set $x=0$: $f(0)=0$ The $y$-intercept is $(0,0)$.

Vertex: $x\text{-coordinate}=\frac{0+4}{2}=2$ Substituting $x=2$ into the function: $f(2)=4(2{)}^2-4(2)=16-8=8$ The vertex is $(2,8)$. Since $a>0$, the parabola opens upwards and is symmetric about $x=2$.

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Part (vi)

Given the function: $f(x)=-x^2-x+6$

Here $a=-1,b=-1,c=6$. To find the factors, we set $f(x)=0$: $\begin{array}{rl}6-x^2-x& =0\\ -(x^2+x-6)& =0\\ x^2+3x-2x-6& =0\\ x(x+3)-2(x+3)& =0\\ (x+3)(x-2)& =0\end{array}$ So, $x=-3$ or $x=2$. The graph intersects the $x$-axis at $(-3,0)$ and $(2,0)$.

To find the $y$-intercept, set $x=0$: $f(0)=6$ The $y$-intercept is $(0,6)$.

Vertex: $x\text{-coordinate}=\frac{-3+2}{2}=-0.5$ Substituting $x=-0.5$ into the function: $f(-0.5)=6-(-0.5{)}^2-(-0.5)=5.25$ The vertex is $(-0.5,5.25)$. Since $a<0$, the parabola opens downwards and is symmetric about $x=-0.5$.

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Key Formulas or Methods Used

• Factoring Quadratic Equations: Solving $a{x}^2+bx+c=0$ to find $x$-intercepts.
• $y$-intercept: Calculated by evaluating $f(0)$.
• Vertex $x$-coordinate: $x_v=\frac{x_1+x_2}{2}$ (midpoint of roots).
• Parabola Direction: If $a>0$, opens upwards; if $a<0$, opens downwards.

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Summary of Steps

1. Identify coefficients $a,b,$ and $c$ from the quadratic function.
2. Factor the equation $f(x)=0$ to find the $x$-intercepts.
3. Find the $y$-intercept by calculating $f(0)$.
4. Determine the vertex $x$-coordinate by finding the midpoint between the $x$-intercepts.
5. Calculate the vertex $y$-coordinate by plugging the $x$-coordinate back into the original function.
6. Determine the opening direction based on the sign of $a$ and sketch the graph.