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how do you find the x-intercepts and the vertex of this parabola: y= -x²-5x ?

Sagot :

konrad509
[tex]-x^2-5x=0\\ -x(x+5)=0\\ x=0 \vee x=-5[/tex]

The x-intercepts are [tex](0,0)[/tex] and [tex](-5,0)[/tex].

[tex]\hbox{vertex}=(h,k)\\ h=\frac{x_1+x_2}{2},k=y(h)\\\\ h=\frac{0-5}{2}=\frac{-5}{2}=-2.5\\ k=y(-2.5)=-(-2.5)^2-5\cdot(-2.5)\\ k=-6.25+12.5\\ k=6.25\\\\ \hbox{vertex}=(2.5,6.25)[/tex]

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