How do you use the important points to sketch the graph of #y=x^2+6x+5#?

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Answer 1 · 2016-03-24T16:01:48

The vertex V of this parabola is #(-3, -4)#. Axis is x - #-3#,. upwards in y-direction. It cuts x-axis at #A(1/2, 0)#and B#(-7/2, 0)#.

The equation can be rearranged as #(x+3)^2=4(1/4)(y+4)#.

This represents a parabola with vertex at #(-3, -4)#.

The parameter for size a = #1/4#.

The curve is above this vertex and is symmetrical about its axis #x=-3#. This is parallel to y-axis.

Put y= o.and solve #x^2+6x+5=0#

#x=-1/2 and -7/2#. So, the curve cuts x-axis at #(1/2, 0), (-7/2, 0)#

The focus #(-3, -15/4)# is right above vertex at a distance #1/4#.

The directrix #y--17/4) is equidistant below the vertex and is parallel to the x-axis..

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