Question

What is the vertex of the parabola with the equation y=15x2+2x−8?

Answer 1

Explanation:

`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=-(b)/(2a)\\\mathrm{The\:parabola\:params\:are:}\\a=15,\:b=2,\:c=-8\\x_v=-(b)/(2a)\\x_v=-(2)/(2\cdot \:15)\\\mathrm{Simplify\:}-(2)/(2\cdot \:15):\quad -(1)/(15)\\x_v=-(1)/(15)\\\mathrm{Plug\:in}\:\:x_v=-(1)/(15)\:\mathrm{to\:find\:the}\:y_v\:\mathrm{value}\\y_v=-(121)/(15)\\`

`\mathrm{Therefore\:the\:parabola\:vertex\:is}\\\left(-(1)/(15),\:-(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(-(1)/(15),\:-(121)/(15)\right)`