How many real solutions does f (x) = 2x21 have, based on the graph?

Question

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(1 point)

two real solutions because the graph is symmetric about the y-axis

no real solutions because the graphs neither touches nor crosses the x-axis

one real solution because the graph intersects the y-axis once

one real solution because the graph touches y-axis at y = -1

Answer 1

To determine how many real solutions the function \( f(x) = 2x^2 + 1 \) has, we can analyze its form and characteristics.

1. The function is a quadratic function of the form \( ax^2 + bx + c \) with \( a = 2 \), \( b = 0 \), and \( c = 1 \).

2. The coefficient \( a \) is positive, meaning the graph opens upward.

3. To find the x-intercepts (real solutions), we need to solve the equation \( 2x^2 + 1 = 0 \). Rearranging gives:

   \[ 2x^2 = -1 \] \[ x^2 = -\frac{1}{2} \]

Since \( x^2 \) cannot be negative for real numbers, this means there are no real solutions.

Based on this analysis, the correct answer is:

no real solutions because the graph neither touches nor crosses the x-axis.