 Graph the given​ function, ​(b) find all values of x where the function is​ discontinuous, and ​(c) find the limit from the left and the right at any values of x where the function is discontinuous.
​h(x)equals
StartSet Start 2 By 2 Matrix 1st Row 1st Column 5 x plus 3 2nd Column if x less than or equals 0 2nd Row 1st Column x squared minus 3 x plus 3 2nd Column if x greater than 0 EndMatrix
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Part 1
​(a) Choose the correct graph below.

The given function \( h(x) \) is a piecewise function defined as follows:

\[
h(x) =
\begin{cases} 
5x + 3 & \text{if } x \leq 0 \\
x^2 - 3x + 3 & \text{if } x > 0
\end{cases}
\]

### Part (a): Graph the function
The function consists of two parts:
1. \( 5x + 3 \) when \( x \leq 0 \)
2. \( x^2 - 3x + 3 \) when \( x > 0 \)

We can break this down:
- The part \( 5x + 3 \) is a straight line with slope 5 and intercept at \( (0, 3) \). Since it is defined for \( x \leq 0 \), we will draw this line for \( x \leq 0 \).
- The part \( x^2 - 3x + 3 \) is a parabola, which opens upwards because the leading coefficient (of \( x^2 \)) is positive. We will plot this curve for \( x > 0 \).

Now, we can plot these two pieces accordingly.

### Part (b): Find all values of \( x \) where the function is discontinuous
To determine if the function is discontinuous at any points, we check the behavior of the function at \( x = 0 \), since the function changes from one piece to another at that point.

For \( x = 0 \):
- The value from the left (using \( 5x + 3 \)) is \( 5(0) + 3 = 3 \).
- The value from the right (using \( x^2 - 3x + 3 \)) is \( 0^2 - 3(0) + 3 = 3 \).

Since the left-hand and right-hand values at \( x = 0 \) are the same and the function is defined at \( x = 0 \), the function is **continuous** at \( x = 0 \).

Thus, there are **no discontinuities** for this function.

### Part (c): Find the limits from the left and right at any points of discontinuity
Since we found no discontinuities in part (b), there are no limits to calculate at points of discontinuity.

The limits at \( x = 0 \) are as follows:
- Left-hand limit: \( \lim_{x \to 0^-} h(x) = 3 \)
- Right-hand limit: \( \lim_{x \to 0^+} h(x) = 3 \)

Both limits agree, so the function is continuous at \( x = 0 \).

Would you like me to generate the graph for the function and explain anything further?

Graph the given function, (b) find all values of x where the function is discontinuous, and (c) find the limit from the left and the right at any values of x where the function is discontinuous. The function is defined as: h(x) = { 5x + 3 if x ≤ 0; x^2 - 3x + 3 if x > 0 }

This problem focuses on graphing a piecewise function, analyzing its continuity, and calculating limits at points of potential discontinuity. Learn how to evaluate limits and determine continuity for piecewise-defined functions. Suitable for students in Grades 10-12.

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