Answered

Westonci.ca is the premier destination for reliable answers to your questions, brought to you by a community of experts. Join our Q&A platform to connect with experts dedicated to providing precise answers to your questions in different areas. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.

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

Sagot :

[tex]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}[/tex]
Thank you for choosing our service. We're dedicated to providing the best answers for all your questions. Visit us again. Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. Westonci.ca is here to provide the answers you seek. Return often for more expert solutions.