Let \[f(x) =
\begin{cases}
2x^2 - 3&\text{if } x\le 2, \\
ax + 4 &\text{if } x>2.
\end{cases}
\]Find $a$ if the graph of $y=f(x)$ is continuous (which means the graph can be drawn without lifting your pencil from the paper).
If the graph of $f$ is continuous, then the graphs of the two cases must meet when $x=2,$ which (loosely speaking) is the dividing point between the two cases. Therefore, we must have $2\cdot 2^2 -3 = 2a + 4.$ Solving this equation gives $a = \boxed{\frac{1}{2}}.$
\frac{1}{2}
Level 5
Algebra
Problem: Let \[f(x) =
\begin{cases}
2x^2 - 3&\text{if } x\le 2, \\
ax + 4 &\text{if } x>2.
\end{cases}
\]Find $a$ if the graph of $y=f(x)$ is continuous (which means the graph can be drawn without lifting your pencil from the paper).
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To ensure the function \( f(x) \) is continuous at \( x = 2 \), the values of the two pieces of the function must be equal at that point. This means we need to evaluate \( f(2) \) from both cases of the piecewise function.

First, calculate \( f(2) \) using the first case where \( x \le 2 \):
\[ f(2) = 2(2^2) - 3 = 2(4) - 3 = 8 - 3 = 5. \]

Next, we need to find the value of \( f(x) \) as \( x \) approaches 2 from the right, using the second case where \( x > 2 \):
\[ f(2) = a(2) + 4 = 2a + 4. \]

For continuity at \( x = 2 \), we set the two expressions equal to each other:
\[ 5 = 2a + 4. \]

Now, solve for \( a \):
1. Subtract 4 from both sides:
   \[ 5 - 4 = 2a \]
   \[ 1 = 2a. \]
2. Divide both sides by 2:
   \[ a = \frac{1}{2}. \]

Thus, the value of \( a \) that ensures continuity of the function at \( x = 2 \) is \( \frac{1}{2} \).<|END_OF_LATENT|> The final answer is $\frac{1}{2}$.