Question

What is the range of this? h(x) = 2x^2+x -1?

Answer 1

\[[-1,+ \infty)\]

Answer 2

range is, by convention, determined by the highest and lowest points; and excluding what nneds to be excluded

Answer 3

since this is an equation of a parabola that opens upward; we only need to determine the vertex position which will be the lowest point

Answer 4

the vertex can be gotten from the determinant of the quadratic formula: -b/2a will be the x part

Answer 5

ax^2 +bx+c is the general quadratic; from this we see that -1/4 is the determinant of your equation

Answer 6

2(-1/4)^2 -(1/4)-1 = -9/8 for the lowest y value in the range

Answer 7

therefore our range is [-9/8,inf)

Answer 8

y of the vertex is =-D/4a isn't it?

Answer 9

y is determined by plugging "D" back into the equation and solving

Answer 10

-1/4
----- not = -9/8 so id venture to say no :)
4 (2)

Answer 11

D=b^2-4ac=9 tht's wht i am talking abt

Answer 12

r-(-9/8)