Question

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

• la règle de la fonction f est de la forme f(x) = a(x − 36)² + k.
• f(36) = 12
• f(0) = −15
• un des zéros de la fonction f est aussi lun des zéros de la fonction g.
• la table de valeurs suivante représente la fonction g.

x	g(x)
120	0

• le point v est le sommet de la parabole représentant la fonction g.

quelles sont les coordonnées du point v ?

Explanation:

Step1: Determine vertex of \( f(x) \)

Given \( f(x) = a(x - 36)^2 + k \), the vertex is \( (36, k) \). Since \( f(36) = 12 \), then \( k = 12 \). So \( f(x) = a(x - 36)^2 + 12 \).

Step2: Find \( a \) using \( f(0) = -15 \)

Substitute \( x = 0 \), \( f(0) = a(0 - 36)^2 + 12 = -15 \).
\( 1296a + 12 = -15 \)
\( 1296a = -27 \)
\( a = \frac{-27}{1296} = -\frac{1}{48} \).
Thus, \( f(x) = -\frac{1}{48}(x - 36)^2 + 12 \).

Step3: Find zeros of \( f(x) \)

Set \( f(x) = 0 \):
\( -\frac{1}{48}(x - 36)^2 + 12 = 0 \)
\( (x - 36)^2 = 12 \times 48 = 576 \)
\( x - 36 = \pm 24 \)
\( x = 36 + 24 = 60 \) or \( x = 36 - 24 = 12 \).
So zeros of \( f(x) \) are \( x = 12 \) and \( x = 60 \).

Step4: Determine zero of \( g(x) \) (common with \( f(x) \))

Given \( g(120) = 0 \), so one zero of \( g(x) \) is \( 120 \). The other zero must be common with \( f(x) \), so either \( 12 \) or \( 60 \). Let's check \( g(0) = -120 \).

Step5: Find vertex of \( g(x) \) (parabola)

For a parabola, the vertex \( x \)-coordinate is the midpoint of the two zeros. Let the zeros of \( g(x) \) be \( x_1 \) and \( x_2 \). We know \( x_2 = 120 \), and \( x_1 \) is \( 12 \) or \( 60 \).

Case 1: \( x_1 = 12 \), \( x_2 = 120 \). Midpoint: \( \frac{12 + 120}{2} = 66 \).
Case 2: \( x_1 = 60 \), \( x_2 = 120 \). Midpoint: \( \frac{60 + 120}{2} = 90 \).

Now, let's verify with \( g(0) = -120 \). Let \( g(x) = b(x - x_1)(x - x_2) \).

Case 1: \( g(x) = b(x - 12)(x - 120) \). At \( x = 0 \): \( g(0) = b(-12)(-120) = 1440b = -120 \) ⇒ \( b = -\frac{120}{1440} = -\frac{1}{12} \).

Case 2: \( g(x) = b(x - 60)(x - 120) \). At \( x = 0 \): \( g(0) = b(-60)(-120) = 7200b = -120 \) ⇒ \( b = -\frac{120}{7200} = -\frac{1}{60} \).

Now, check the vertex form. For a parabola \( g(x) = b(x - h)^2 + k \), vertex is \( (h, k) \).

For Case 2: \( g(x) = -\frac{1}{60}(x - 90)^2 + k \). At \( x = 120 \), \( g(120) = 0 \):
\( -\frac{1}{60}(120 - 90)^2 + k = 0 \)
\( -\frac{1}{60}(900) + k = 0 \)
\( -15 + k = 0 \) ⇒ \( k = 15 \).

Check \( g(0) \): \( -\frac{1}{60}(0 - 90)^2 + 15 = -\frac{8100}{60} + 15 = -135 + 15 = -120 \), which matches \( g(0) = -120 \).

For Case 1: \( g(x) = -\frac{1}{12}(x - 66)^2 + k \). At \( x = 120 \):
\( -\frac{1}{12}(120 - 66)^2 + k = 0 \)
\( -\frac{1}{12}(2916) + k = 0 \)
\( -243 + k = 0 \) ⇒ \( k = 243 \).
Check \( g(0) \): \( -\frac{1}{12}(0 - 66)^2 + 243 = -\frac{4356}{12} + 243 = -363 + 243 = -120 \), which also matches. Wait, but we need to see the graph. The graph of \( g(x) \) has vertex V, and \( f(x) \) and \( g(x) \) intersect. Wait, maybe I made a mistake. Wait, the zeros of \( f(x) \) are 12 and 60, and \( g(x) \) has a zero at 120. Let's re-examine the graph: both parabolas open downward (since \( f(x) \) has \( a = -1/48 < 0 \), and \( g(x) \) also opens downward as per the graph).

Wait, the key is: the two parabolas intersect, and V is the vertex of \( g(x) \). Let's re-express \( g(x) \). Since \( g(120) = 0 \) and let the other zero be \( x = 60 \) (since 60 is between 0 and 120, more likely). So zeros at 60 and 120. Then vertex x-coordinate is \( \frac{60 + 120}{2} = 90 \). Then, \( g(x) = b(x - 60)(x - 120) \). At \( x = 0 \), \( g(0) = b(-60)(-120) = 7200b = -120 \) ⇒ \( b = -120 / 7200 = -1/60 \). Then, \( g(x) = -\frac{1}{60}(x - 60)(x - 120) \). To find the vertex, expand \( g(x) \):
\( g(x) = -\frac{1}{60}(x^2 - 180x + 7200) = -\frac{1}{60}x^2 + 3x - 120 \).
The vertex x-coordinate is \( -\frac{b}{2a} = -\frac{3}{2 \times (-\frac{1}{60})} = 90 \). Then, \( g(90) = -\frac{1}{60}(90 - 60)(90 - 120) = -\frac{1}{60}(30)(-30) = 15 \). So vertex V is \( (90, 15) \).

Wait, but let's confirm with \( f(x) \) and \( g(x) \) intersection. But the key steps: find \( f(x) \)'s zeros, determine common zero with \( g(x) \), find vertex of \( g(x) \) as midpoint of its zeros, then confirm with \( g(0) \). The correct vertex is \( (90, 15) \) because when zeros are 60 and 120, midpoint is 90, and \( g(90) = 15 \) matches the calculation.

Answer:

\( (90, 15) \)