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Sagot :
ANSWER
The vertex is (-7, 4)
EXPLANATION
We usually see the equation of a parabola with this form:
[tex]y=a(x-h)^2+k[/tex]Where (h, k) is the vertex of the parabola.
But in this problem, the variables are changed, so we're looking for an equation like:
[tex]x=b(y-k)^2+h[/tex]The vertex is also point (h, k).
To complete the square we have to see the terms that contain y: y² - 8y
We know that a perfect square has the form:
[tex](a\pm b)^2=a^2\pm2ab+b^2[/tex]If the term that contains y with exponent 1 has a coefficient 8, then we can find the second number of the binomial for the perfect square:
[tex](y+n)^2=y^2+2yn+\cdots=y^2-8y+\cdots[/tex]let's not think of the third term just yet. For now we have to concentrate on the second term:
[tex]\begin{gathered} 2yn=-8y \\ 2n=-8 \\ n=-4 \end{gathered}[/tex]The second number for the perfect square is -4:
[tex](y-4)^2=y^2-8y+16[/tex]Now we have to put this inside the equation. Since now we have a third term that is 16 we have to subtract 16 so we keep the equation true. I'll replace y²-8y by (y-4)²-16:
[tex]-4x+(y-4)^2-16-12=0[/tex]We can add 4x on both sides so we have the x and y separate:
[tex]\begin{gathered} 4x=(y-4)^2-16-12 \\ 4x=(y-4)^2-28 \end{gathered}[/tex]And finally we have to divide both sides by 4:
[tex]x=\frac{1}{4}(y-4)^2-7[/tex]So in this case k = 4 and h = -7. The vertex is (-7, 4)
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