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)
0	−120
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 k for f(x)

Given \( f(x) = a(x - 36)^2 + k \) and \( f(36) = 12 \). Substitute \( x = 36 \):
\( f(36) = a(36 - 36)^2 + k = k = 12 \). So \( k = 12 \), and \( f(x) = a(x - 36)^2 + 12 \).

Step2: Find a for f(x)

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

Step3: Find zeros of f(x)

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

Step4: Identify common zero with g(x)

Given \( g(120) = 0 \), so \( x = 120 \) is a zero of \( g \). The other zero of \( g \) must be the common zero with \( f \), so either \( x = 12 \) or \( x = 60 \).

Step5: Determine vertex of g(x)

A parabola \( g(x) \) has vertex at the midpoint of its zeros. Let the zeros of \( g \) be \( x_1 \) and \( x_2 \). We know one zero is \( x = 120 \), and the other is \( x = 60 \) (since \( x = 12 \) would make midpoint \( \frac{12 + 120}{2} = 66 \), but let's verify with \( x = 60 \)):
Midpoint (x-coordinate of vertex V) is \( \frac{60 + 120}{2} = 90 \).

Step6: Find y-coordinate of V

First, find the equation of \( g(x) \). Let \( g(x) = m(x - 60)(x - 120) \). Use \( g(0) = -120 \):
\( -120 = m(0 - 60)(0 - 120) \)
\( -120 = 7200m \)
\( m = -\frac{120}{7200} = -\frac{1}{60} \).
Thus, \( g(x) = -\frac{1}{60}(x - 60)(x - 120) \).
Expand or use vertex form: The vertex x-coordinate is 90. Substitute \( x = 90 \) into \( g(x) \):
\( g(90) = -\frac{1}{60}(90 - 60)(90 - 120) \)
\( = -\frac{1}{60}(30)(-30) \)
\( = -\frac{1}{60}(-900) = 15 \).

Answer:

Les coordonnées du point V sont \( \boldsymbol{(90, 15)} \).