How many real solutions does g(x)=−1.6−0.2x2

Question

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.

Answer 1

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.