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ANSWER
[tex](-1.24,0),(3.24,0)[/tex]EXPLANATION
We want to find the x intercepts of the parabola with vertex (1, 20) and y intercept (0, 16).
To do this, first find the equation of the parabola.
The vertex form of the equation of a parabola is:
[tex]\begin{gathered} y=a(x-h)^2+k \\ \text{where (h, k) = vertex} \end{gathered}[/tex]Therefore, we have:
[tex]y=a(x-1)^2+20[/tex]To find the value of a, substitute the values of x and y of the y intercept into the equation:
[tex]\begin{gathered} 16=a(0-1)^2+20 \\ 16=a(-1)^2+20 \\ 16=a+20 \\ \Rightarrow a=16-20 \\ a=-4 \end{gathered}[/tex]Therefore, the equation of the parabola in vertex form is:
[tex]y=-4(x-1)^2+20[/tex]Now, write it in standard form:
[tex]\begin{gathered} y=-4(x^2-2x+1)+20 \\ y=-4x^2+8x-4+20 \\ y=-4x^2+8x+16 \end{gathered}[/tex]To find the x intercepts, find the values of x when y = 0:
[tex]\begin{gathered} -4x^2+8x+16=0 \\ \text{Solve using Quadratic formula:} \\ x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ a=-4;b=8;c=16 \\ x=\frac{-8\pm\sqrt[]{8^2-4(-4)(16)}}{2(-4)}=\frac{-8\pm\sqrt[]{64+256}}{-8} \\ x=\frac{-8+\sqrt[]{320}}{-8};x=\frac{-8-\sqrt[]{320}}{-8} \\ \Rightarrow x=\frac{-8+17.89}{-8};x=\frac{-8-17.89}{-8} \\ x=-1.24;x=3.24 \end{gathered}[/tex]Therefore, the x intercepts are:
[tex](-1.24,0),(3.24,0)[/tex]
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