EXPLANATION
We first can draw the vertex in order to obtain the minimum value:
[tex]\mathrm{The\:vertex\:of\:an\:up-down\:facing\:parabola\:of\:the\:form}\:y=ax^2+bx+c\:\mathrm{is}\:x_v=-\frac{b}{2a}[/tex][tex]\mathrm{The\:parabola\:params\:are:}[/tex][tex]a=1,\:b=-7,\:c=10[/tex][tex]x_v=-\frac{b}{2a}[/tex][tex]x_v=-\frac{\left(-7\right)}{2\cdot \:1}[/tex]
Simplify:
[tex]y_v=-\frac{9}{4}[/tex][tex]\mathrm{Therefore\:the\:parabola\:vertex\:is}[/tex][tex]\left(\frac{7}{2},\:-\frac{9}{4}\right)[/tex]
Now, we need to compute the intercepts:
Plug y=0 into the equation and solve the resulting equation 0=x^2-7x+10:
The x-intercept are the following:
(2,0) and (5,0)
Identifying the y-intercept:
Plug x=0 into the equation and solve the resulting equation y=10 for y:
[tex]y=0^2-7*0+10[/tex]
Computing the power and multiplying terms:
[tex]y=10[/tex]
The y-intercept is at (0,10)
In conclusin, the appropriate option is OPTION B
