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

Question

anonymous · Answer

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

amistre64 · Answer

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

amistre64 · Answer

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

amistre64 · Answer

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

amistre64 · Answer

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

amistre64 · Answer

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

amistre64 · Answer

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

angela210793 · Answer

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

amistre64 · Answer

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

amistre64 · Answer

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

angela210793 · Answer

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

anonymous · Answer

r-(-9/8)