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Sagot :
[tex]y=x^2-8x+12[/tex]
To calculate the x-intercepts we replace y=0 and solve for x
[tex]x^2-8x+12=0[/tex]where a is 1, b is -8 and c 12
so factor the expression using
[tex]x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]replacing
[tex]\begin{gathered} x=\frac{-(-8)\pm\sqrt[]{(-8)^2-4(1)(12)^{}}}{2(1)} \\ \\ x=\frac{8\pm\sqrt[]{64-48}}{2} \\ \\ x=\frac{8\pm\sqrt[]{16}}{2} \\ \\ x=\frac{8\pm4}{2} \\ \\ x=4\pm2 \end{gathered}[/tex]so x have two solutions because there are two x-intercepts
[tex]\begin{gathered} x_1=4+2=6 \\ x_2=4-2=2 \end{gathered}[/tex]then the x-intercepts are 6 and 2, the corrdinates are
[tex]\begin{gathered} (6,0) \\ (2,0) \end{gathered}[/tex]Vertex
the vertex is a point (x,y) to calculate x we use
[tex]x=\frac{-b}{2a}[/tex]and replace
[tex]\begin{gathered} x=\frac{-(-8)}{2(1)} \\ \\ x=\frac{8}{2}=4 \end{gathered}[/tex]now replace x=4 on the equation of the parable to find y
[tex]\begin{gathered} y=x^2-8x+12 \\ y=(4)^2-8(4)+12 \\ y=16-32+12 \\ y=-4 \end{gathered}[/tex]the coordinate of the vertex is
[tex](4,-4)[/tex]
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