Question

write a function in any form that would match the graph shown below.
(graph with x-axis from -10 to 10, y-axis from -500 to 500, curve crossing x-axis at -8, touching x-axis at 5 (or around), y-intercept at 150 (approx), and end behavior increasing on right, decreasing then increasing. answer section: f(x) = blank submit answer. you have up to 7 questions left to raise your...

Explanation:

Step1: Identify roots and multiplicity

The graph crosses the x - axis at \(x = - 8\) (single root, so factor \((x + 8)\)) and touches the x - axis at \(x = 5\) (double root, so factor \((x - 5)^2\)). So the polynomial has factors \((x + 8)\) and \((x - 5)^2\). Let the function be \(f(x)=a(x + 8)(x - 5)^2\).

Step2: Find the leading coefficient \(a\)

Use the y - intercept. The graph passes through \((0,150)\) (wait, looking at the graph, when \(x = 0\), \(y=150\)? Wait, no, the graph at \(x = 0\) is at \(y = 150\)? Wait, the grid: from the graph, when \(x = 0\), the y - value is 150? Wait, let's re - check. The graph at \(x = 0\) is at \(y = 150\)? Wait, the y - axis has 100, 200, etc. Wait, the point at \(x = 0\) is \((0,150)\)? Wait, no, the graph at \(x = 0\) is at \(y = 150\)? Wait, let's use the point \((0,150)\) (assuming). Substitute \(x = 0\), \(y = 150\) into \(f(x)=a(x + 8)(x - 5)^2\):

\(150=a(0 + 8)(0 - 5)^2\)

\(150=a\times8\times25\)

\(150 = 200a\)

\(a=\frac{150}{200}=\frac{3}{4}\)? Wait, no, maybe I misread the y - intercept. Wait, the graph at \(x = 0\) is at \(y = 150\)? Wait, looking at the graph again, the y - intercept is at \(y = 150\)? Wait, no, the grid lines: each square is, say, 50? Wait, no, the graph at \(x = 0\) is at \(y = 150\)? Wait, maybe the y - intercept is \(y = 150\). Wait, but let's recalculate. Wait, \((0 - 5)^2=25\), \(0 + 8 = 8\), so \(8\times25 = 200\). So \(a=\frac{y}{200}\). If \(y = 150\), then \(a=\frac{150}{200}=\frac{3}{4}\). But wait, maybe the y - intercept is \(y = 150\). Wait, but let's check the graph again. Alternatively, maybe the y - intercept is \(y = 150\). So \(f(x)=\frac{3}{4}(x + 8)(x - 5)^2\). Wait, but let's expand \((x + 8)(x - 5)^2=(x + 8)(x^{2}-10x + 25)=x^{3}-10x^{2}+25x+8x^{2}-80x + 200=x^{3}-2x^{2}-55x + 200\). Then \(f(x)=\frac{3}{4}(x^{3}-2x^{2}-55x + 200)=\frac{3}{4}x^{3}-\frac{3}{2}x^{2}-\frac{165}{4}x + 150\). But maybe there is a miscalculation. Wait, maybe the y - intercept is \(y = 150\). Alternatively, maybe the leading coefficient is \(\frac{3}{4}\). But let's check with another approach. Wait, the graph is a cubic function? No, it's a cubic? Wait, no, the graph has a double root at \(x = 5\) and a single root at \(x=-8\), so it's a cubic function (degree 3). Wait, but the graph's shape: when \(x\to\infty\), \(y\to\infty\), and when \(x\to-\infty\), \(y\to-\infty\), which is consistent with a cubic with positive leading coefficient.

Wait, maybe I made a mistake in the y - intercept. Let's look at the graph again. The graph at \(x = 0\) is at \(y = 150\)? Wait, the y - axis: the point is at \(y = 150\). So substituting \(x = 0\) into \(f(x)=a(x + 8)(x - 5)^2\):

\(f(0)=a(8)(25)=200a\). If \(f(0)=150\), then \(a=\frac{150}{200}=\frac{3}{4}\). So the function is \(f(x)=\frac{3}{4}(x + 8)(x - 5)^2\).

Wait, but let's expand it:

\(f(x)=\frac{3}{4}(x + 8)(x^{2}-10x + 25)=\frac{3}{4}(x^{3}-10x^{2}+25x+8x^{2}-80x + 200)=\frac{3}{4}(x^{3}-2x^{2}-55x + 200)=\frac{3}{4}x^{3}-\frac{3}{2}x^{2}-\frac{165}{4}x + 150\)

Alternatively, maybe the y - intercept is \(y = 150\), so this function works.

Answer:

\(f(x)=\frac{3}{4}(x + 8)(x - 5)^2\) (or expanded form \(\frac{3}{4}x^{3}-\frac{3}{2}x^{2}-\frac{165}{4}x + 150\))