y=-3x^2+6x-2
Sketch the graph of the given porabola and state the coordinance of its intercepts.
I can graph it if I can get some help with the vertex and coordinance.
This is what I got; vertex:(1,1) x-intercept:(-1.6,.-4), and y-intercept:(0,-2). Is this right?
4 answers
that's suppose to be -.4
what is thr perimeter of an equilateral triangle that measurea 4 in one side
Rachel: 12. Perimeter means to "walk around" the entire object. If the side of an equilateral triangle is 4, so are the other 2 sides (4 + 4 + 4).
No, Jennifer, you are close.
How can you have an x-intercept of (-1.6, -.4) ??
x-intercepts have the form (?,0)
Your vertex is correct, and your y-intercept is correct
to find the x-intercepts, let y = 0
0 = -3x^2+6x-2
3x^2 - 6x + 2 = 0
x = (6 ± √12)/6
= 1.577 or .423
those are the x values of the intercepts
so the x-intercepts are
(1.6,0) and (.4,0)
(you actually had the opposite of these numbers, so you must have made a sign error.
Also notice that for any parabola the midpoint of the x-intercepts gives you the x value of the vertex. That would not happen for your values, but
(1.577 + .423)/2 = 1, the x value of the vertex)
How can you have an x-intercept of (-1.6, -.4) ??
x-intercepts have the form (?,0)
Your vertex is correct, and your y-intercept is correct
to find the x-intercepts, let y = 0
0 = -3x^2+6x-2
3x^2 - 6x + 2 = 0
x = (6 ± √12)/6
= 1.577 or .423
those are the x values of the intercepts
so the x-intercepts are
(1.6,0) and (.4,0)
(you actually had the opposite of these numbers, so you must have made a sign error.
Also notice that for any parabola the midpoint of the x-intercepts gives you the x value of the vertex. That would not happen for your values, but
(1.577 + .423)/2 = 1, the x value of the vertex)
0 answers