my vertex is (-6,-36)

my domain all real numbers yx²+12x

my range is y≥-36

how to graph

my vertex is (-6,-36)

my domain all real numbers yx²+12x

my range is y≥-36

how to graph

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I'm assuming that the equation y=x^{2}+12x (the equal sign isn't shown in your question). You can make a sketch very quickly and easily. Because it is a positive quadratic equation, it will go up in the positive and negative x directions (looks like a smile). You have one point at x=-6 and y=-36. This is the bottom of the curve.

If you need the sketch to be more accurate then you can include the zeroes. You can factor the equation: y=x(x+12). This means that when y=0, x=0 and x+12=0. Meaning: x=0,-12. This gives you two more points to graph: (0,0) and (-12,0).

Because you are in precalculus, your teacher may also want you to show the focus and directrix. If not, ignore this part, the above graph should be sufficient. The easiest way to find the focus is to put the equation in vertex form. You do this by completing the square. y+36=x^{2}+12x+36 (half of the x term squared added to both sides). y+36=(x+6)^{2}. The general form for a quadratic in vertex form is y-k=4p(x-h)^{2}. The p value lets you find the focus. In this case: 4p=1, so p=1/4. This means that the focus is 1/4 away from your vertex. Because it is a positive quadratic, the focus is above the vertex. This means the focus is at (-6,-35.75). The directrix is a horizontal line p away from the vertex in the opposite direction as the focus. The directrix is at y=-36.25. Hope this helped!