Question

4x^2-5x-15=0 how to find the vertex and y intercept

Answer 1

`ax^2+bx+c=0\\\\the\ vertex:\left((-b)/(2a);(-(b^2-4ac))/(4a)\right)\\\\y-intrcept=c\\=======================================\\\\4x^2-5x-15=0\\\\a=4;\ b=-5;\ c=-15\\\\(-b)/(2a)=(-(-5))/(2\cdot4)=(5)/(8);\ (-(b^2-4ac))/(4a)=(-[(-5)^2-4\cdot4\cdot(-15)])/(4\cdot4)=-(265)/(16)\\\\\boxed{the\ vertex:\left((5)/(8);-(265)/(16)\right)}\\\\\boxed{y-intercept:-15}`