The table of values represents points on the graph of a quadratic function.
x -2 -1 0 1 2
y 31 16 7 4 7
I got Vertex (1,4) & f(x) = -7x^2 + 14x - 3 as my equation (standard form).
May someone check my work please
I like the vertex, but clearly the graph opens upward, so you cannot have -7x^2
show your work, and I think you will spot your own mistake. Start with
y = a(x-1)^2 + 4
and just find a
On second try I got y= 3x^2 -6x + 7
Does that look good?
looks better. I assume you checked to make sure the points fit.
To check if your work is correct, we can use the values from the table to determine if they satisfy the equation you found.
The equation you derived, in standard form, is:
f(x) = -7x^2 + 14x - 3
Now, let's substitute the x-values from the table into the equation and see if we get the corresponding y-values:
For x = -2:
f(-2) = -7(-2)^2 + 14(-2) - 3
= -7(4) - 28 - 3
= -28 - 28 - 3
= -59
The y-value for x = -2 in the table is 31, so the equation doesn't match. Thus, (1, 4) is not the vertex.
For x = -1:
f(-1) = -7(-1)^2 + 14(-1) - 3
= -7(1) - 14 - 3
= -7 - 14 - 3
= -24
The y-value for x = -1 in the table is 16, so the equation doesn't match.
For x = 0:
f(0) = -7(0)^2 + 14(0) - 3
= 0 + 0 - 3
= -3
The y-value for x = 0 in the table is 7, so the equation doesn't match.
For x = 1:
f(1) = -7(1)^2 + 14(1) - 3
= -7(1) + 14 - 3
= -7 + 14 - 3
= 4
The y-value for x = 1 in the table is 4, so the equation matches.
For x = 2:
f(2) = -7(2)^2 + 14(2) - 3
= -7(4) + 28 - 3
= -28 + 28 - 3
= -3
The y-value for x = 2 in the table is 7, so the equation matches.
Since the equation matches for x = 1 and x = 2, it seems like the equation you found is correct. However, (1, 4) is not the vertex of the quadratic function corresponding to the given table. You may want to recheck your calculations for the vertex.