Question

PLEASE HELP I NEED THIS ASAP
Find the vertex of y=x^2-7x+4

Answer 1

`\mathrm{Minimum}\space\left((7)/(2),\:-(33)/(4)\right)`

Explanation:

`y=x^2-7x+4\\\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=1,\:b=-7,\:c=4\\x_v=-(b)/(2a)\\x_v=-(\left(-7\right))/(2\cdot \:1)\\\mathrm{Simplify}\\x_v=(7)/(2)\\\mathrm{Plug\:in}\:\:x_v=(7)/(2)\:\mathrm{to\:find\:the}\:y_v\:\mathrm{value}\\y_v=\left((7)/(2)\right)^2-7\cdot (7)/(2)+4\\`

`\mathrm{Simplify\:}\left((7)/(2)\right)^2-7\cdot (7)/(2)+4:\quad -(33)/(4)\\y_v=-(33)/(4)\\Therefore\:the\:parabola\:vertex\:is\\\left((7)/(2),\:-(33)/(4)\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=1\\\mathrm{Minimum}\space\left((7)/(2),\:-(33)/(4)\right)`