Question

situation dapplication : trois fonctions polynomiales
considérons les fonctions polynomiales du second degré f, g et h représentées ci-
dessous dans le plan cartésien.

• la règle de la fonction f est f(x) = -0,25(x + 10)(x + 2).
• la règle de la fonction g est g(x) = -0,5x² + 10x - 50.
• le sommet de la fonction f est lun des points de la parabole représentant la

fonction h.

• un des zéros de la fonction f est lun des zéros de la fonction h.
• le sommet de la fonction g est lun des points de la parabole représentant la

fonction h.

quelle est la règle de la fonction h ?

Explanation:

Answer:

To determine the rule of function \( h \), we analyze the given information about functions \( f \) and \( g \), and use the properties of quadratic functions (vertex, zeros) to find the equation of \( h \).

Step 1: Find the zeros of \( f(x) \)

The function \( f(x) = -0.25(x + 10)(x + 2) \) is in factored form. To find its zeros, set \( f(x) = 0 \):

$$ -0.25(x + 10)(x + 2) = 0 $$

Since \( -0.25
eq 0 \), we solve \( (x + 10)(x + 2) = 0 \). This gives \( x = -10 \) or \( x = -2 \). So the zeros of \( f \) are \( x = -10 \) and \( x = -2 \).

Step 2: Find the vertex of \( f(x) \)

For a quadratic function in factored form \( f(x) = a(x - r_1)(x - r_2) \), the \( x \)-coordinate of the vertex is the midpoint of the zeros:

$$ x_{\text{vertex}} = \frac{r_1 + r_2}{2} $$

For \( f(x) = -0.25(x + 10)(x + 2) \), the zeros are \( r_1 = -10 \) and \( r_2 = -2 \). Thus:

$$ x_{\text{vertex}} = \frac{-10 + (-2)}{2} = \frac{-12}{2} = -6 $$

To find the \( y \)-coordinate, substitute \( x = -6 \) into \( f(x) \):

$$ f(-6) = -0.25(-6 + 10)(-6 + 2) = -0.25(4)(-4) = -0.25(-16) = 4 $$

So the vertex of \( f \) is \( (-6, 4) \).

Step 3: Find the vertex of \( g(x) \)

The function \( g(x) = -0.5x^2 + 10x - 50 \) is in standard form \( ax^2 + bx + c \). The \( x \)-coordinate of the vertex is given by \( x = -\frac{b}{2a} \):

$$ x = -\frac{10}{2(-0.5)} = -\frac{10}{-1} = 10 $$

To find the \( y \)-coordinate, substitute \( x = 10 \) into \( g(x) \):

$$ g(10) = -0.5(10)^2 + 10(10) - 50 = -0.5(100) + 100 - 50 = -50 + 100 - 50 = 0 $$

So the vertex of \( g \) is \( (10, 0) \).

Step 4: Determine the zeros and vertex of \( h(x) \)

From the problem:

• One zero of \( f \) is a zero of \( h \). Let's use \( x = -2 \) (or \( x = -10 \); we'll verify later).
• The vertex of \( f \) \( (-6, 4) \) is on \( h \).
• The vertex of \( g \) \( (10, 0) \) is on \( h \). Also, since \( g(10) = 0 \), \( (10, 0) \) is a zero of \( h \) (because it's on the \( x \)-axis).

So \( h(x) \) has a zero at \( x = 10 \) (from \( g \)'s vertex) and let's confirm the other zero. We know one zero of \( f \) is a zero of \( h \). Let's check if \( x = -2 \) or \( x = -10 \) works with the vertex of \( f \) and \( g \)'s vertex.

Wait, the vertex of \( g \) is \( (10, 0) \), so \( (10, 0) \) is a zero of \( h \). Let's assume the other zero is \( x = -2 \) (from \( f \)'s zero). Now, we have two points on \( h \): \( (-6, 4) \) (vertex of \( f \)), \( (10, 0) \) (vertex of \( g \) and zero of \( h \)), and \( (-2, 0) \) (zero of \( f \) and zero of \( h \))? Wait, no—let's re-examine.

Wait, the vertex of \( f \) is \( (-6, 4) \), which is on \( h \). The vertex of \( g \) is \( (10, 0) \), which is on \( h \) and is a zero (since \( y = 0 \)). Also, one zero of \( f \) is a zero of \( h \). Let's check the graph: the parabola \( h \) opens upwards (since it's a "U" shape). Let's list the known points on \( h \):

• Zero at \( x = 10 \) (from \( g \)'s vertex, \( (10, 0) \))
• Zero at \( x = -2 \) (from \( f \)'s zero, let's verify if \( (-2, 0) \) is on \( h \))
• Vertex at \( (-6, 4) \)? Wait, no—wait, the vertex of \( f \) is \( (-6, 4) \), which is on \( h \). Let's check the parabola: if \( h \) has zeros at \( x = -2 \) and \( x = 10 \), then its axis of symmetry is the midpoint of \( -2 \) and \( 10 \):

$$ x = \frac{-2 + 10}{2} = \frac{8}{2} = 4 $$

But the vertex of \( f \) is \( (-6, 4) \), which would not be on a parabola with axis of symmetry \( x = 4 \). So we must have made a mistake. Let's re-express:

Wait, the vertex of \( g \) is \( (10, 0) \), which is a zero (since \( y = 0 \)), so \( (10, 0) \) is a root. The vertex of \( f \) is \( (-6, 4) \), which is on \( h \). Also, one root of \( f \) is a root of \( h \). Let's take the root of \( f \) as \( x = -2 \) (so \( (-2, 0) \) is a root) and \( x = 10 \) as the other root. Then the axis of symmetry is \( x = \frac{-2 + 10}{2} = 4 \). The vertex would be at \( (4, k) \). But \( (-6, 4) \) is on \( h \), so let's check:

If \( h(x) = a(x + 2)(x - 10) \), then substitute \( (-6, 4) \):

$$ 4 = a(-6 + 2)(-6 - 10) = a(-4)(-16) = 64a \implies a = \frac{4}{64} = \frac{1}{16} $$

But let's check the vertex of \( g \): \( (10, 0) \) is a root, so that's good. Now check the vertex of \( f \): \( (-6, 4) \) is on \( h \), so that's good. Now, let's verify with the vertex of \( g \): \( (10, 0) \) is on \( h \), which is correct.

Wait, but let's check the other root. If \( h(x) = \frac{1}{16}(x + 2)(x - 10) \), let's expand it:

$$ h(x) = \frac{1}{16}(x^2 - 10x + 2x - 20) = \frac{1}{16}(x^2 - 8x - 20) = \frac{1}{16}x^2 - \frac{1}{2}x - \frac{5}{4} $$

But let's check the vertex of \( h \). For \( h(x) = \frac{1}{16}(x + 2)(x - 10) \), the vertex is at \( x = 4 \) (midpoint of \( -2 \) and \( 10 \)). Substitute \( x = 4 \):

$$ h(4) = \frac{1}{16}(6)(-6) = \frac{1}{16}(-36) = -\frac{9}{4} = -2.25 $$

But the vertex of \( f \) is \( (-6, 4) \), which is not the vertex of \( h \), so our assumption of the roots is wrong.

Wait, let's re-express. The vertex of \( f \) is \( (-6, 4) \), which is on \( h \). The vertex of \( g \) is \( (10, 0) \), which is on \( h \). Also, one root of \( f \) is a root of \( h \). Let's find the roots of \( f \): \( x = -10 \) and \( x = -2 \). Let's check if \( (10, 0) \) is a root, and another root is \( x = -10 \). Then the axis of symmetry is \( x = \frac{-10 + 10}{2} = 0 \). The vertex would be at \( (0, k) \). But \( (-6, 4) \) is on \( h \), so:

\( h(x) = a(x + 10)(x - 10) \) (since roots at \( -10 \) and \( 10 \)). Then \( h(-6) = a(-6 + 10)(-6 - 10) = a(4)(-16) = -64a = 4 \implies a = -\frac{4}{64} = -\frac{1}{16} \). But then \( h(10) = 0 \), which is correct, but \( h(-6) = 4 \), which is correct. Wait, but the vertex of \( g \) is \( (10, 0) \), which is a root, and the vertex of \( f \) is \( (-6, 4) \), which is on \( h \). Let's check the graph: \( h \) would open downward (since \( a = -\frac{1}{16} < 0 \)), but the graph of \( h \) in the diagram is a "U" shape (opening upward). So that's a contradiction.

Wait, the graph of \( h \) in the diagram: looking at the sketch, \( h \) is a parabola opening upwards (since it has a minimum). So \( a > 0 \). Let's re-express:

We know:

• \( h \) passes through \( (-6, 4) \) (vertex of \( f \))
• \( h \) passes through \( (10, 0) \) (vertex of \( g \), which is on the \( x \)-axis, so a root)
• \( h \) has another root, which is a root of \( f \): either \( x = -10 \) or \( x = -2 \)

Let's assume the roots are \( x = -2 \) and \( x = 10 \). Then the equation is \( h(x) = a(x + 2)(x - 10) \). Now, substitute \( (-6, 4) \):

\( 4 = a(-6 + 2)(-6 - 10) = a(-4)(-16) = 64a \implies a = \frac{4}{64} = \frac{1}{16} \)

Now, check the vertex of \( h \): for \( h(x) = \frac{1}{16}(x + 2)(x - 10) \), the axis of symmetry is \( x = \frac{-2 + 10}{2} = 4 \). The vertex is at \( (4, h(4)) \). Let's compute \( h(4) \):

\( h(4) = \frac{1}{16}(4 + 2)(4 - 10) = \frac{1}{16}(6)(-6) = \frac{1}{16}(-36) = -\frac{9}{4} = -2.25 \). But \( (-6, 4) \) is on \( h \), which is not the vertex, so that's okay (the vertex of \( f \) is just a point on \( h \), not necessarily the vertex of \( h \)).

Wait, but the problem states: "Le sommet de la fonction \( f \) est l’un des points de la parabole représentant la fonction \( h \)" (The vertex of \( f \) is one of the points of \( h \)) and "Le sommet de la fonction \( g \) est l’un des points de la parabole représentant la fonction \( h \)" (The vertex of \( g \) is one of the points of \( h \)). Also, "Un des zéros de la fonction \( f \) est l’un des zéros de la fonction \( h \)" (One zero of \( f \) is a zero of \( h \)).

From the graph, \( h \) intersects \( f \) at two points: one is the vertex of \( f \)? No, the graph shows \( h \) intersecting \( f \) at a point, and \( h \) intersecting \( g \) at a point (the vertex of \( g \), which is on the \( x \)-axis).

Wait, let's re-express all known points:

• \( f \)'s vertex: \( (-6, 4) \in h \)
• \( g \)'s vertex: \( (10, 0) \in h \)
• One zero of \( f \): \( x = -2 \) or \( x = -10 \in h \) (on \( x \)-axis)

Looking at the graph, \( h \) has a minimum (opens upward), intersects \( f \) (which opens downward) at two points: one is the vertex of \( f \)? No, the vertex of \( f \) is a peak, and \( h \) is a valley. Wait, the graph: \( f \) is a downward parabola, \( g \) is a downward parabola, \( h \) is an upward parabola.

So \( h \) (upward) intersects \( f \) (downward) at two points: one is the vertex of \( f \)? No, the vertex of \( f \) is a point on \( h \), and another intersection point. Also, \( h \) intersects \( g \) (downward) at the vertex of \( g \) (which is on the \( x \)-axis) and another point.

Wait, let's use the two roots and the vertex. Let's suppose \( h \) has roots at \( x = -2 \) and \( x = 10 \), and passes through \( (-6, 4) \). Then:

\( h(x) = a(x + 2)(x - 10) \)

Substitute \( (-6, 4) \):

\( 4 = a(-6 + 2)(-6 - 10) = a(-4)(-16) = 64a \implies a = \frac{4}{64} = \frac{1}{16} \)

So \( h(x) = \frac{1}{16}(x + 2)(x - 10) \)

Let's expand this:

\( h(x) = \frac{1}{16}(x^2 - 10x + 2x - 20) = \frac{1}{16}(x^2 - 8x - 20) = \frac{1}{16}x^2 - \frac{1}{2}x - \frac{5}{4} \)

Now, check if \( (10, 0) \) is on \( h \): \( h(10) = \frac{1}{16}(10 + 2)(10 - 10) = 0 \), correct.

Check if \( (-2, 0) \) is on \( h \): \( h(-2) = \frac{1}{16}(-2 + 2)(-2 - 10) = 0 \), correct.

Check if \( (-6, 4) \) is on \( h \): \( h(-6) = \frac{1}{16}(-6 + 2)(-6 - 10) = \frac{1}{16}(-4)(-16) = \frac{1}{16}(64) = 4 \), correct.

Now, check the vertex of \( h \): for \( h(x) = \frac{1}{16}(x + 2)(x - 10) \), the axis of symmetry is \( x = \frac{-2 + 10}{2} = 4 \). The vertex is at \( (4, h(4)) \):

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