Question

What is the domain and range of the quadratic equation y = -3(x - 10)2 + 5?

Answer 1

can anyone help me?!

Answer 2

domain of any polynomial, unless otherwise stated, is all real numbers

Answer 3

Domain: All Real Numbers
Range: y greater than or equal to 5
Domain: All Real Numbers
Range: y less than or greater to 5
Domain: All Real Numbers
Range: y greater than or equal to 10
Domain: All Real Numbers
Range: y less than or greater to 10

these are the answer choices..

Answer 4

as for the range, since this is a parabola that opens down (you can tell by the \(-3\) out front) it will go from \(-\infty\) up to the maximum value, which evidently is \(5\) because of the \(+5\) out at the end

Answer 5

so which one is it exactly?:/

Answer 6

btw there is no such thing as "less than or greater to 5 "

Answer 7

answer please?

Answer 8

i wrote the answer above

Answer 9

still confused

Answer 10

Domain for all polynomials is all real x.
Range for a quadratic will be greater than or equal the minimum or less than or equal to the maximum point.
y = -3(x - 10)2 + 5?
This is an example of a concave down quadratic (since coefficient of x^2 is negative).
So it has a maximum turning point. This occurs for x = -b/2a in general form. If you expand - 3(x-10)2 = -3(x^2 - 20x + 100) = -3x^2 +60x + 100, the coefficient of x^2 is -3 and of x is 60 so x = -b/2a = -60/2(-3) = 10. Now we need to substitute this value in to find y at the maximum. y = -3(10 - 10)^2 + 5 = -3(0)^2 + 5 = 5.
So y < or = 5 is the range.
alternatively, y = -3(x - 10)2 + 5? is just y = y = -3(x - 10)2 shifted up 5 units, so the maximum of y = -3(x - 10)2 is shifted 5 units up. The maximum value of y = -3(x - 10)2 is easily obtained --> it is a perfect square so its maximum occurs at y = 0 (the x-axis). Hence we shift 5 units up from 0 to get y = 5 being the maximum. The range is all y below and inclusive of this maximum.