Answer:
Step-by-step explanation:
[tex]y=15x^2+2x-8\\\mathrm{Parabola\:equation\:in\:polynomial\:form}\\\mathrm{The\:vertex\:of\:an\:up-down\:facing\:parabola\:of\:the\:form}\:y=ax^2+bx+c\:\mathrm{is}\:x_v=-\frac{b}{2a}\\\mathrm{The\:parabola\:params\:are:}\\a=15,\:b=2,\:c=-8\\x_v=-\frac{b}{2a}\\x_v=-\frac{2}{2\cdot \:15}\\\mathrm{Simplify\:}-\frac{2}{2\cdot \:15}:\quad -\frac{1}{15}\\x_v=-\frac{1}{15}\\\mathrm{Plug\:in}\:\:x_v=-\frac{1}{15}\:\mathrm{to\:find\:the}\:y_v\:\mathrm{value}\\y_v=-\frac{121}{15}\\[/tex]
[tex]\mathrm{Therefore\:the\:parabola\:vertex\:is}\\\left(-\frac{1}{15},\:-\frac{121}{15}\right)\\\mathrm{If}\:a<0,\:\mathrm{then\:the\:vertex\:is\:a\:maximum\:value}\\\mathrm{If}\:a>0,\:\mathrm{then\:the\:vertex\:is\:a\:minimum\:value}\\a=15\\\mathrm{Minimum}\space\left(-\frac{1}{15},\:-\frac{121}{15}\right)[/tex]
