Question

Compare the graphs of f(x)=2 square root x+1 and g(x)=2 3root x+1 . What is the y-intercept of both graphs?

Answer 1

To compare the graphs of the functions \( f(x) = 2\sqrt{x} + 1 \) and \( g(x) = 2\sqrt[3]{x} + 1 \), we can start by determining the y-intercepts of both functions.

The y-intercept of a function is found by evaluating the function at \( x = 0 \).

1. **For the function \( f(x) = 2\sqrt{x} + 1 \)**:
\[
f(0) = 2\sqrt{0} + 1 = 2 \cdot 0 + 1 = 1
\]
The y-intercept of \( f(x) \) is \( (0, 1) \).

2. **For the function \( g(x) = 2\sqrt[3]{x} + 1 \)**:
\[
g(0) = 2\sqrt[3]{0} + 1 = 2 \cdot 0 + 1 = 1
\]
The y-intercept of \( g(x) \) is also \( (0, 1) \).

Thus, both functions \( f(x) \) and \( g(x) \) have the same y-intercept at the point \( (0, 1) \).

### Graph Characteristics:
- **Domain**:
- \( f(x) = 2\sqrt{x} + 1 \) is defined for \( x \geq 0 \) (since square root is only defined for non-negative values).
- \( g(x) = 2\sqrt[3]{x} + 1 \) is defined for all real numbers \( x \).

- **Range**:
- The range of \( f(x) \) is \( [1, \infty) \) because \( \sqrt{x} \) starts from 0, meaning \( f(x) \) starts from 1.
- The range of \( g(x) \) is \( (-\infty, \infty) \) since \( \sqrt[3]{x} \) can take any real value.

- **End Behavior**:
- As \( x \) approaches infinity, \( f(x) \) will also approach infinity.
- For \( g(x) \), as \( x \) approaches positive infinity, \( g(x) \) will also approach infinity, and as \( x \) approaches negative infinity, \( g(x) \) will approach negative infinity.

- **Graph Shape**:
- The graph of \( f(x) \) will have a more pronounced curve and only exists in the first quadrant since \( f(x) \) is only defined for \( x \geq 0 \).
- The graph of \( g(x) \) is S-shaped, passing through the point (0, 1) and extending to negative values for negative \( x \).

In summary, both functions share the y-intercept at \( (0, 1) \) but differ in domain, range, and overall shape of their graphs.