Question

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

Answer 1

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