How do you use the important points to sketch the graph of y=x^2+6x+5y=x2+6x+5?

1 Answer
A. S. Adikesavan
Mar 24, 2016

The vertex V of this parabola is (-3, -4)(−3,−4). Axis is x - -3−3,. upwards in y-direction. It cuts x-axis at A(1/2, 0)A(12,0)and B(-7/2, 0)(−72,0).

Explanation:

The equation can be rearranged as (x+3)^2=4(1/4)(y+4)(x+3)2=4(14)(y+4).

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

The parameter for size a = 1/414.

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

Put y= o.and solve x^2+6x+5=0x2+6x+5=0

x=-1/2 and -7/2x=−12and−72. So, the curve cuts x-axis at (1/2, 0), (-7/2, 0)(12,0),(−72,0)

The focus (-3, -15/4)(−3,−154) is right above vertex at a distance 1/414.

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

.

Related questions
See all questions in Quadratic Functions and Their Graphs
Impact of this question
1473 views around the world
You can reuse this answer
Creative Commons License