I have to tell you that
NONE of the options are appropriate.
You can check it by substituting several "observable" points on the parabola e.g. (-1, 1.5) and (-1.5, 2.5) to get the y-coordinate.
The correct answer is x^2+0.5x+1.
Here displays the full solution to the question:
From the graph, you should identify the following:
1) The graph is a parabola (major axis // y-axis).
⇒ It represents a quadratic function.
∴ G, H eliminated
*2) The graph does not intersect the x-axis.
⇒ △<0
∴ A, B, F eliminated
*3) y-intercept of the graph = 1
⇒ The constant term is 1. (no elimination)
*Step 2, 3 may be skipped for FASTER calculation of the answer.
Note that when the graph is translated along the -ve y-axis by 1 unit, there will be two distinct real roots of 0 and -0.5.
Ax^2+Bx+C = A(x-root1)(x-root2)
y-1 = A(x-0)(x+0.5) where A is a real number
y = A(x^2+0.5x)+1 ...(1)
To find out A, substitute a point on the parabola (-1, 1.5) into (1): (x^2+0.5x≠0)
1.5 = A[(-1)^2+0.5(-1)]+1
1.5 = 0.5A+1
A = 1
∴The equation is x^2+0.5x+1.
The nearest but INCORRECT answer is E, 4x^2+2x+1.
If we compare it with equation (1), A=4, so the roots at (-0.5, 1) and (0, 1) will return a positive test result.
However, when we test the point (-1, 1.5) with y=4x^2+2x+1,
we will get
LHS = 1.5
RHS = 4(-1)^2+2(-1)+1
= 4-2+1
= 3 ≠ LHS
The graphs are plotted as below:
http://xs132.xs.to/xs132/08434/vhgraph001519.gif