How do you find the vertex, the y-intercept, and symmetric point, and use these to sketch the graph given #y=x^2+8x+21#?

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Answer 1 · 2016-11-20T13:36:03

See the answer below

Let's write the equation in the standard vertex form.

#y=x^2+8x+21=x^2+8x+16+5#

#y=(x+4)^2+5#

The vertex is #(-4,5)#

The symmetric line is #x=-4#

When #x=0#, #=>#, #y=21#

The intercept with the y axis is #(0,21)#

There are no intercept with the x-axis.

graph{x^2+8x+21 [-30.5, 27.22, -2, 26.88]}