Question

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

Answer 1

`-x^2-5x=0\\ -x(x+5)=0\\ x=0 \vee x=-5`

The x-intercepts are
`(0,0)` and
`(-5,0)`.


`\hbox{vertex}=(h,k)\\ h=(x_1+x_2)/(2),k=y(h)\\\\ h=(0-5)/(2)=(-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)`