Given: #color(blue)(y=x^2+10x+21)#

#color(brown)("To find the vertex")#

Consider the +10 from +10x

#color(brown)(x_("vertex")) = (-1/2) xx (+10)color(brown)(=-5)#

Substitute the found value of #x_("vertex")# in the original equation to find #y_("vertex")#

#y_("vertex")=(-5)^2+10(-5)+21#

#color(brown)(y_("vertex")=-4#

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Each intercept is when the plotted line crosses one of the axis. That is: when #x=0# the plot crosses #y_("intercept")# and when #y=0# the plot crosses #x_("intercept")#

#color(brown)("So to find "y_("intercept") " write the equation as:")#

#y_("intercept")=0^2+10(0)+21#

#color(brown)(y_("intercept")=+21)#

You will observe from this that it is the value of the constant.

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#color(brown)("So to find "x_("intercept") " write the equation as:")#

#0=x^2+10x+21#

I notice that #3xx7=21# and #3+7=10#

so we can factorise giving:

#color(blue)(0=(x+3)(x+7))color(green)( -> x^2+7x+3x+21) = x^2+10x+21#

So we have to make each of the bracketed parts 0 to give #y=0#

So for #(x+3)=0# the value of #x=-3#

and for #(x+7)=0# the value of #x=-7#

So #color(brown)(x_("intercept") = -7 " and " -3)#