Given the function:
[tex]f\mleft(x\mright)=2\mleft(x-5\mright)^2-3[/tex]
You can identify that it is a parabola because it is a Quadratic Function.
By definition, the Parent Function (the simplest form) of Quadratic Functions, is:
[tex]f(x)=x^2[/tex]
And its graph is:
Notice that its vertex is at the Origin.
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In this case, you have the Quadratic Function written in Vertex Form:
[tex]f(x)=a(x-h)^2+k[/tex]
Where "h" is the x-coordinate of the vertex (it indicates the horizontal shift) and "k" is the y-coordinate of the vertex (it indicates the vertical shift). The value of "a" indicates if the parabola is stretched or compressed:
- If:
[tex]|a|<1[/tex]
It is compressed.
- If:
[tex]|a|>1[/tex]
It is stretched.
- If "a" is negative, the parabola opens downward.
- If "a" is positive, the parabola opens upward.
In this case, you can identify that:
[tex]\begin{gathered} a=2 \\ h=5 \\ k=-3 \end{gathered}[/tex]
You can find the x-intercepts as follows:
1. Make:
[tex]f(x)=0[/tex]
2. Solve for "x".
Then, you get:
[tex]\begin{gathered} 0=2\mleft(x-5\mright)^2-3 \\ \\ \frac{3}{2}=(x-5)^2 \end{gathered}[/tex][tex]\begin{gathered} \sqrt[]{\frac{3}{2}}=\sqrt[]{(x-5)^2} \\ \\ \pm\sqrt[]{\frac{3}{2}}=x-5 \\ \\ 5\pm\sqrt[]{\frac{3}{2}}=x\Rightarrow\begin{cases}x_1=5+\sqrt[]{\frac{3}{2}}\approx6.225 \\ \\ x_2=5-\sqrt[]{\frac{3}{2}}\approx3.775\end{cases} \end{gathered}[/tex]
Knowing all the data, you can graph the parabola.
Hence, the answer is:
- Graph:
- Vertical shift of 3 units down:
[tex]k=-3[/tex]
- Horizontal shift of 5 units to the right:
[tex]h=5[/tex]
- Vertical stretch by a factor of 2:
[tex]a=2[/tex]
