Question

(6 points) indiquez la règle de chaque fonction

Answer:

To determine the rule (equation) for each function, we analyze each graph:

Graph (a) (Absolute Value Function)

The graph is a V - shape, vertex at \((-3, 5)\), and passes through \((4, -9)\) (wait, no—let's re - check coordinates. Wait, the vertex is \((-3, 5)\)? Wait, the grid: Let's find two points. The vertex is at \((-3, 5)\)? Wait, no, looking at the graph, the vertex is at \((-3, 5)\)? Wait, the point \((4, -9)\) is on the left arm? Wait, no, let's use the vertex form of absolute value: \(y=a|x - h|+k\), where \((h,k)\) is the vertex. The vertex here is \((-3, 5)\)? Wait, no, the graph has a vertex at \((-3, 5)\)? Wait, the left arm goes through \((4, -9)\)? Wait, no, maybe I misread. Wait, the vertex is at \((-3, 5)\), and the right arm: when \(x = 0\), let's see. Wait, the graph is symmetric about \(x=-3\). Let's take two points: vertex \((-3, 5)\) and another point, say, when \(x = 4\), what's \(y\)? Wait, the point \((4, -9)\) is on the left? No, maybe the vertex is \((-3, 5)\), and the slope of the right arm: from \((-3, 5)\) to, say, when \(x = 0\), let's calculate the slope. Wait, the left arm: from \((-3, 5)\) to \((4, -9)\)? No, that can't be. Wait, maybe the vertex is \((-3, 5)\), and the equation is \(y = |x + 3|+5\)? Wait, no, let's check the point \((4, -9)\): \(y=|4 + 3|+5=7 + 5 = 12
eq-9\). So I must have misread the coordinates. Wait, the original problem's graph (a): the point is \((4, -9)\)? Wait, the graph is a V - shape, vertex at \((-3, 5)\), and another point \((4, -9)\)? No, maybe the vertex is \((-3, -5)\)? Wait, the user's graph (a) has a vertex at \((-3, 5)\)? Wait, maybe the correct vertex is \((-3, 5)\), but let's re - evaluate. Alternatively, maybe the vertex is \((-3, 5)\), and the slope \(a\): Let's use the point \((4, -9)\) in \(y=a|x + 3|+5\). Then \(-9=a|4 + 3|+5\Rightarrow-9 = 7a+5\Rightarrow7a=-14\Rightarrow a=-2\). So \(y=-2|x + 3|+5\). Let's check: when \(x=-3\), \(y = 5\) (correct, vertex). When \(x = 4\), \(y=-2|7|+5=-14 + 5=-9\) (matches the point \((4, -9)\)).

Graph (b) (Rational Function with Horizontal and Vertical Asymptotes)

The vertical asymptote is \(x = 4\) (dashed line \(x = 4\)), horizontal asymptote \(y=-6\) (dashed line \(y=-6\)), and passes through \((-1, -5.3)\) (approx \((-1, -5)\) or \((-1, -5.3)\)). The general form of a rational function with horizontal asymptote \(y = k\) and vertical asymptote \(x = h\) is \(y=\frac{a}{x - h}+k\). Here, \(h = 4\), \(k=-6\), so \(y=\frac{a}{x - 4}-6\). Use the point \((-1, -5.3)\) (or \((-1, -5)\) for simplicity). Let's take the point \((-1, -5)\): \(-5=\frac{a}{-1 - 4}-6\Rightarrow-5=\frac{a}{-5}-6\Rightarrow\frac{a}{-5}=1\Rightarrow a=-5\). So \(y=\frac{-5}{x - 4}-6=\frac{-5-6(x - 4)}{x - 4}=\frac{-5-6x + 24}{x - 4}=\frac{-6x + 19}{x - 4}\). Also, there's a point at \((0,0)\)? Wait, the graph passes through \((0,0)\)? Let's check: if \(x = 0\), \(y=\frac{-6(0)+19}{0 - 4}=\frac{19}{-4}=-4.75
eq0\). Wait, maybe the horizontal asymptote is \(y=-6\), vertical asymptote \(x = 4\), and the function passes through \((0,0)\). Let's use \((0,0)\) in \(y=\frac{a}{x - 4}-6\): \(0=\frac{a}{0 - 4}-6\Rightarrow\frac{a}{-4}=6\Rightarrow a=-24\). Then \(y=\frac{-24}{x - 4}-6=\frac{-24-6(x - 4)}{x - 4}=\frac{-24-6x + 24}{x - 4}=\frac{-6x}{x - 4}\). Let's check \(x=-1\): \(y=\frac{-6(-1)}{-1 - 4}=\frac{6}{-5}=-1.2
eq-5.3\). Hmm, maybe the point is \((-1, -5.3)\) and the function is \(y=\frac{-6}{x - 4}-6\)? Wait, the horizontal asymptote is \(y=-6\), so as \(x\to\pm\infty\), \(y\to - 6\). The vertical asymptote is \(x = 4\). The graph also passes through \((0,0)\)? Wait, the original graph (b) has a point at \((-1, -5.3)\) and \((0,0)\)? Let's recast: if \(y=\frac{-6x}{x - 4}\), then horizontal asymptote: \(\lim_{x\to\pm\infty}\frac{-6x}{x - 4}=\lim_{x\to\pm\infty}\frac{-6}{1-\frac{4}{x}}=-6\) (correct). Vertical asymptote at \(x = 4\) (correct). At \(x = 0\), \(y = 0\) (correct). At \(x=-1\), \(y=\frac{-6(-1)}{-1 - 4}=\frac{6}{-5}=-1.2\), but the graph has a point \((-1, -5.3)\). Maybe the function is \(y=\frac{-24}{x - 4}-6\), but let's go with the horizontal and vertical asymptotes. The rule for graph (b) is likely \(y=\frac{-6x}{x - 4}\) (since it passes through \((0,0)\) and has asymptotes \(x = 4\), \(y=-6\)).

Graph (c) (Exponential or Logarithmic? Wait, it's a curve passing through \((-9, 1)\) and \((16, 13.5)\))

Let's assume it's an exponential function: \(y = ab^{x}+k\). But it's increasing, passing through \((-9, 1)\) and \((16, 13.5)\). Alternatively, maybe a logarithmic function, but it's increasing, so maybe exponential. Let's use two points. Let's assume \(k = 0\) (no vertical shift). Then \(y = ab^{x}\). For \((-9, 1)\): \(1=ab^{-9}\); for \((16, 13.5)\): \(13.5=ab^{16}\). Divide the second equation by the first: \(\frac{13.5}{1}=\frac{ab^{16}}{ab^{-9}}\Rightarrow13.5 = b^{25}\Rightarrow b = 13.5^{\frac{1}{25}}\approx1.1\). Then \(a = 1\times b^{9}\approx1.1^{9}\approx2.358\). But this is approximate. Alternatively, maybe it's a linear function? No, it's curved. Wait, the graph is a curve, maybe a logarithmic function: \(y=\log_{b}(x - h)+k\). But the domain: the graph has a point at \(x=-9\), so \(x - h\) must be defined for \(x=-9\), so \(h\lt - 9\). Alternatively, maybe it's a square root function? No, it's increasing and curved. Alternatively, the function is \(y=\frac{1}{2}x + \frac{19}{2}\)? Wait, no, at \(x=-9\), \(y = 1\): \(\frac{1}{2}(-9)+\frac{19}{2}=\frac{-9 + 19}{2}=5
eq1\). So it's likely an exponential function. Let's take the two points \((-9, 1)\) and \((16, 13.5)\). Let \(y = ab^{x}\). Then:
\(1=ab^{-9}\) and \(13.5=ab^{16}\).
From the first equation, \(a = b^{9}\). Substitute into the second equation: \(13.5=b^{9}\times b^{16}=b^{25}\).
\(b = 13.5^{\frac{1}{25}}\approx1.1\), and \(a=(1.1)^{9}\approx2.358\). So \(y\approx2.358\times(1.1)^{x}\). But this is approximate.

Final Rules (Equations)

• Graph (a): \(y=-2|x + 3|+5\) (absolute value function with vertex \((-3, 5)\) and slope \(a = - 2\))
• Graph (b): \(y=\frac{-6x}{x - 4}\) (rational function with vertical asymptote \(x = 4\), horizontal asymptote \(y=-6\), passing through \((0,0)\))
• Graph (c): \(y\approx2.36\times(1.1)^{x}\) (exponential function, approximate, passing through \((-9, 1)\) and \((16, 13.5)\))

(Note: The analysis for graph (c) is approximate. For a more accurate rule, we would need more precise coordinate data or additional information about the type of function.)