Look at the axis graph. If both of the coordinates are negative, it would be in quadrant lll. If you go over to the x-axis, you would be in quadrant lV, or 4.
(-2,-3) -> (2,-3)
You change the x-factor's sign.
I hope this helps!
~kaikers
Answer:
(d) f(x) = -x²
Step-by-step explanation:
For the vertex of the quadratic function to be at the origin, both the x-term and the constant must be zero. That is, the function must be of the form ...
f(x) = a(x -h)² +k . . . . . . . . . . vertex form; vertex at (h, k)
f(x) = a(x -0)² +0 = ax² . . . . . vertex at the origin, (h, k) = (0, 0)
Of the offered answer choices, the only one with a vertex at the origin is ...
f(x) = -x² . . . . . a=-1
12+11x=17x11x+12=17x
(both sides of the equation)
11x+12−17x=17x−17x
(Subtract 17x from both sides)
−6x+12=0−6x+12−12=0−12
(Subtract 12 from both sides)
−6x=−12−6x−6=−12−6
(Divide both sides by -6)
x=2
Check answers. (Plug them in to make sure they work.)
x=2
(Works in original equation)
