Question

complex numbers review
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the curve passes through the following approximate coordinates: left parenthesis negative 5 comma negative 10 right parenthesis, left parenthesis negative 4 comma 0 right parenthesis, peaks at left parenthesis negative 2.5 comma 5 right parenthesis, descends to a valley at left parenthesis negative 0.4 comma 3 right parenthesis, then rises to left parenthesis 0.4 comma 4 right parenthesis, left parenthesis 1 comma 6 right parenthesis, and left parenthesis 1.5 comma 10 right parenthesis
which of the following equations would give the correct number of real and non - real solutions to match the number of x - intercepts shown on the graph?
(1 point)
○ $y=(x + 3)(x + 4)$
○ $y=(x^{2}-2)(x + 4)$
○ $y=(x^{2}+2)(x + 4)$
○ $y=(x - 4)$

Explanation:

Step1: Analyze x - intercepts from graph

From the graph description, the curve passes through \((-4,0)\) and we need to check other x - intercepts. Let's analyze each equation's x - intercepts (where \(y = 0\)) and the nature of roots (real/non - real).

Step2: Analyze Option 1: \(y=(x + 3)(x + 4)\)

Set \(y = 0\), then \((x + 3)(x + 4)=0\). The solutions are \(x=-3\) and \(x=-4\). But from the graph, we need to check if the number of x - intercepts and non - real roots match. This is a quadratic (degree 2) with two real roots. But let's check other options.

Step3: Analyze Option 2: \(y=(x^{2}-2)(x + 4)\)

Set \(y = 0\), then \(x^{2}-2=0\) or \(x + 4=0\). For \(x^{2}-2=0\), \(x=\pm\sqrt{2}\) (real roots), and for \(x + 4=0\), \(x=-4\) (real root). So total real roots: 3. But from the graph, let's see the behavior. Wait, the graph has one x - intercept at \(x=-4\) (from the point \((-4,0)\))? Wait, no, the graph description: the curve passes through \((-4,0)\), and let's check the other points. Wait, maybe I misread. Wait, the graph's x - intercept: from the points, \((-4,0)\) is one. Let's check the degree and roots.

Step4: Analyze Option 3: \(y=(x^{2}+2)(x + 4)\)

Set \(y = 0\), then \(x^{2}+2=0\) or \(x + 4=0\). For \(x^{2}+2=0\), \(x^{2}=-2\), so \(x=\pm i\sqrt{2}\) (non - real roots). For \(x + 4=0\), \(x=-4\) (real root). So we have 1 real root (\(x=-4\)) and 2 non - real roots. Now check the graph: the curve has one x - intercept at \(x = - 4\) (since it passes through \((-4,0)\)) and the other roots from \(x^{2}+2\) are non - real. Let's check the other options.

Step5: Analyze Option 4: \(y=(x - 4)\)

Set \(y = 0\), then \(x = 4\). But the graph has an x - intercept at \(x=-4\), so this is incorrect.

Now, let's re - check the graph's x - intercepts. The graph has an x - intercept at \(x=-4\) (from \((-4,0)\)). Now, for the equation \(y=(x^{2}+2)(x + 4)\), when \(y = 0\), \(x=-4\) (real) and \(x^{2}+2=0\) (non - real). From the graph, how many x - intercepts? The graph passes through \((-4,0)\), so one x - intercept. Let's check the degree: the equation \(y=(x^{2}+2)(x + 4)=x^{3}+4x^{2}+2x + 8\) (degree 3). The graph's behavior: the curve has a peak and valleys, which is consistent with a cubic function. The equation \(y=(x^{2}-2)(x + 4)\) would have three real roots (\(x=-4,x=\sqrt{2},x =-\sqrt{2}\)), but the graph seems to have only one x - intercept at \(x=-4\) (from the given points: \((-4,0)\) is the only x - intercept? Wait, the graph description: the curve passes through \((-4,0)\), and other points like \((-5,-10)\), peaks at \((-2.5,5)\), etc. So maybe only one x - intercept at \(x=-4\). So for \(y=(x^{2}+2)(x + 4)\), we have one real root (\(x=-4\)) and two non - real roots (from \(x^{2}+2=0\)), which matches the requirement of the number of real and non - real solutions to match the x - intercepts (one x - intercept means one real root, and the other roots are non - real).

Answer:

\(y=(x^{2}+2)(x + 4)\) (the third option: \(y=(x^{2}+2)(x + 4)\))