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