The graph of $y=ax^2+bx+c$ is given below, where $a$, $b$, and $c$ are integers.  Find $a-b+c$.

[asy]
size(150);

Label f;

f.p=fontsize(4);

xaxis(-3,3,Ticks(f, 1.0));

yaxis(-4,4,Ticks(f, 1.0));

real f(real x)

{

return x^2+2x-1;

}

draw(graph(f,-2.7,.7),linewidth(1),Arrows(6));
[/asy]
When $x=-1$, we have $y = a-b+c$.  The graph appears to pass through $(-1,-2)$.  Since $a$, $b$, and $c$ are integers, we know that $y$ is an integer when $x=-1$, so the graph does indeed pass through $(-1,-2)$.  Therefore, $y=-2$ when $x=-1$, so $a-b+c = \boxed{-2}$.
-2
Level 4
Algebra
Problem: The graph of $y=ax^2+bx+c$ is given below, where $a$, $b$, and $c$ are integers.  Find $a-b+c$.

[asy]
size(150);

Label f;

f.p=fontsize(4);

xaxis(-3,3,Ticks(f, 1.0));

yaxis(-4,4,Ticks(f, 1.0));

real f(real x)

{

return x^2+2x-1;

}

draw(graph(f,-2.7,.7),linewidth(1),Arrows(6));
[/asy]
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the values of \(a\), \(b\), and \(c\) from the quadratic equation \(y = ax^2 + bx + c\), we analyze the graph provided. The graph is a parabola that opens upwards, indicating that the coefficient \(a\) is positive. 

Next, we identify the vertex of the parabola. The vertex appears to be at the point \((-1, -2)\). The vertex form of a parabola is given by \(y = a(x-h)^2 + k\), where \((h, k)\) is the vertex. Here, \(h = -1\) and \(k = -2\). 

We can rewrite the equation in vertex form as \(y = a(x + 1)^2 - 2\). To convert this back to standard form, we expand it: 
\[
y = a(x^2 + 2x + 1) - 2 = ax^2 + 2ax + (a - 2).
\]
From this, we can identify \(b = 2a\) and \(c = a - 2\).

Next, we need to find specific integer values for \(a\), \(b\), and \(c\). We can use the point where the graph intersects the y-axis, which occurs at \(x = 0\). Substituting \(x = 0\) into the equation gives us the y-intercept:
\[
y = a(0)^2 + b(0) + c = c.
\]
From the graph, the y-intercept appears to be \(-1\), so we have \(c = -1\).

Now, substituting \(c = -1\) into the equation \(c = a - 2\) gives:
\[
-1 = a - 2 \implies a = 1.
\]

With \(a = 1\), we can find \(b\):
\[
b = 2a = 2(1) = 2.
\]

Now we have \(a = 1\), \(b = 2\), and \(c = -1\). We need to compute \(a - b + c\):
\[
a - b + c = 1 - 2 - 1 = -2.
\]

Thus, the final answer is \(-2\).<|END_OF_LATENT|> The final answer is $-2$.