Question

common core algebra i b - cr
determining the number of x - intercepts
the domain of a quadratic function is all real numbers and the range is ( yleq2 ). how many x - intercepts does the function have?

Explanation:

Step1: Recall quadratic function properties

A quadratic function has the form \( y = ax^2 + bx + c \) (or vertex form \( y = a(x - h)^2 + k \)). The range of a quadratic function depends on the direction the parabola opens (determined by \( a \)) and the vertex. If \( a < 0 \), the parabola opens downward, and the range is \( y \leq k \) (where \( k \) is the y - coordinate of the vertex). If \( a>0 \), it opens upward, and the range is \( y\geq k \).

Step2: Analyze the given range

Here, the range is \( y\leq2 \), so the parabola opens downward (since the maximum value of \( y \) is 2, meaning the vertex is the highest point). The x - intercepts are the points where \( y = 0 \). Since the parabola opens downward and the maximum value of \( y \) is 2 (which is greater than 0), the parabola will cross the x - axis at two points (because the graph goes from \( -\infty \) as \( x\to\pm\infty \) (wait, no, for a downward - opening parabola, as \( x\to\pm\infty \), \( y\to-\infty \)), and since the vertex (maximum point) has \( y = 2>0 \), the graph must cross the x - axis twice. Wait, actually, let's think again. The vertex is at \( y = 2 \) (maximum). Since the parabola opens downward, it will extend downwards from \( y = 2 \). So when \( y = 0 \), we can solve \( ax^2+bx + c=0 \). Since the vertex is above the x - axis ( \( y = 2>0 \)) and the parabola opens downward, the graph will intersect the x - axis at two points. Wait, no, actually, if the vertex is at \( y = k \), and for a downward - opening parabola (\( a<0 \)):

• If \( k>0 \), the parabola intersects the x - axis at two points (because it comes from \( -\infty \) as \( x\to\pm\infty \), reaches a maximum at \( y = k>0 \), then goes back to \( -\infty \), so it must cross \( y = 0 \) twice).
• If \( k = 0 \), it touches the x - axis at one point (vertex on the x - axis).
• If \( k<0 \), it does not intersect the x - axis.

In our case, \( k = 2>0 \) and \( a<0 \) (downward opening), so the number of x - intercepts is 2. Wait, but wait, maybe I made a mistake. Wait, the range is \( y\leq2 \), so the vertex is at \( y = 2 \), and the parabola opens downward. So when we set \( y = 0 \), since \( 0<2 \), and the parabola is opening downward (so it will pass through \( y = 0 \) twice as it goes from \( -\infty \) (as \( x\to\pm\infty \)) up to \( y = 2 \) and then back down). So the number of x - intercepts is 2.

Wait, no, actually, let's take an example. Let the quadratic function be \( y=-x^{2}+2 \). This is a downward - opening parabola with vertex at \( (0,2) \). To find the x - intercepts, set \( y = 0 \):
\( -x^{2}+2=0 \)
\( x^{2}=2 \)
\( x=\pm\sqrt{2} \), which are two distinct real roots. So the number of x - intercepts is 2.

Answer:

2