Asked by yeslian c
How many real solutions does g(x)=−1.6−0.2x2
have, based on its graph?
(1 point)
Responses
The equation g(x)=−1.6−0.2x2
has no real solutions because it has no y-intercepts.
The equation g of x is equal to negative 1 point 6 minus 0 point 2 x squared has no real solutions because it has no y-intercepts. - no response given
The equation g(x)=−1.6−0.2x2
has no real solutions because it has no x-intercepts.
The equation g of x is equal to negative 1 point 6 minus 0 point 2 x squared has no real solutions because it has no x-intercepts. - no response given
The equation g(x)=−1.6−0.2x2
has one real solution because it crosses the y-axis once.
The equation g of x is equal to negative 1 point 6 minus 0 point 2 x squared has one real solution because it crosses the y-axis once. - incorrect
The equation g(x)=−1.6−0.2x2
has no real solutions because it crosses the x-axis once.
have, based on its graph?
(1 point)
Responses
The equation g(x)=−1.6−0.2x2
has no real solutions because it has no y-intercepts.
The equation g of x is equal to negative 1 point 6 minus 0 point 2 x squared has no real solutions because it has no y-intercepts. - no response given
The equation g(x)=−1.6−0.2x2
has no real solutions because it has no x-intercepts.
The equation g of x is equal to negative 1 point 6 minus 0 point 2 x squared has no real solutions because it has no x-intercepts. - no response given
The equation g(x)=−1.6−0.2x2
has one real solution because it crosses the y-axis once.
The equation g of x is equal to negative 1 point 6 minus 0 point 2 x squared has one real solution because it crosses the y-axis once. - incorrect
The equation g(x)=−1.6−0.2x2
has no real solutions because it crosses the x-axis once.
Answers
Answered by
GPT-4o mini
To determine the number of real solutions to the equation \( g(x) = -1.6 - 0.2x^2 \), we first analyze the function itself.
1. **Form of the Function**:
The function \( g(x) = -1.6 - 0.2x^2 \) is a quadratic function, specifically a downward-opening parabola, because the coefficient of \( x^2 \) is negative.
2. **Y-Intercept**:
The y-intercept of the function can be found by evaluating \( g(0) \):
\[
g(0) = -1.6 - 0.2(0)^2 = -1.6.
\]
So the function intersects the y-axis at \( (0, -1.6) \).
3. **X-Intercepts**:
To find the x-intercepts, we set \( g(x) = 0 \):
\[
-1.6 - 0.2x^2 = 0.
\]
Rearranging gives us:
\[
0.2x^2 = -1.6.
\]
This leads to:
\[
x^2 = -\frac{1.6}{0.2} = -8.
\]
Since \( x^2 = -8 \) has no real solutions (as the square of a real number cannot be negative), the function has no x-intercepts.
### Conclusion
The equation \( g(x) = -1.6 - 0.2x^2 \) has **no real solutions** because it has no x-intercepts (the value of \( g(x) \) is always below the x-axis). Therefore, the correct statement is:
**The equation \( g(x) = -1.6 - 0.2x^2 \) has no real solutions because it has no x-intercepts.**
1. **Form of the Function**:
The function \( g(x) = -1.6 - 0.2x^2 \) is a quadratic function, specifically a downward-opening parabola, because the coefficient of \( x^2 \) is negative.
2. **Y-Intercept**:
The y-intercept of the function can be found by evaluating \( g(0) \):
\[
g(0) = -1.6 - 0.2(0)^2 = -1.6.
\]
So the function intersects the y-axis at \( (0, -1.6) \).
3. **X-Intercepts**:
To find the x-intercepts, we set \( g(x) = 0 \):
\[
-1.6 - 0.2x^2 = 0.
\]
Rearranging gives us:
\[
0.2x^2 = -1.6.
\]
This leads to:
\[
x^2 = -\frac{1.6}{0.2} = -8.
\]
Since \( x^2 = -8 \) has no real solutions (as the square of a real number cannot be negative), the function has no x-intercepts.
### Conclusion
The equation \( g(x) = -1.6 - 0.2x^2 \) has **no real solutions** because it has no x-intercepts (the value of \( g(x) \) is always below the x-axis). Therefore, the correct statement is:
**The equation \( g(x) = -1.6 - 0.2x^2 \) has no real solutions because it has no x-intercepts.**