Question

quatre fonctions polynomiales
considérons les fonctions polynomiales du second degré ( f_1, f_2, f_3 ) et ( f_4 ) représentées ci-dessous dans le plan cartésien.

• le point ( v(-48, 40) ) est le sommet de la parabole représentant la fonction ( f_1 ).
• ( f_1(-144) = 200 )
• ( f_1(0) = f_2(0) )
• la règle de la fonction ( f_2 ) est de la forme ( f_2(x) = ax^2 - 2x + c ).
• ( f_2(144) = 224 )
• le point ( s ) est à la fois le sommet de la parabole représentant la fonction ( f_2 ) et le sommet de la parabole représentant la fonction ( f_3 ).
• la règle de la fonction ( f_3 ) est de la forme ( f_3(x) = -\frac{1}{32}(x - h)^2 + k ).
• la fonction ( f_3 ) et la fonction ( f_4 ) ont un zéro commun.
• la table de valeurs suivante représente la fonction ( f_4 ).

( x )	( f_4(x) )
( 0 )	( 16 )

quelle est la règle de la fonction ( f_4 )?

Explanation:

Step1: Find the equation of \( f_1(x) \)

The vertex form of a parabola is \( f(x)=a(x - h)^2 + k \), where \((h,k)\) is the vertex. For \( f_1 \), the vertex \( V(-48,40) \), so \( f_1(x)=a(x + 48)^2+40 \). We know \( f_1(-144) = 200 \), substitute \( x=-144 \) into the equation:
\[ \begin{align*} 200&=a(-144 + 48)^2+40\\ 200 - 40&=a(-96)^2\\ 160&=9216a\\ a&=\frac{160}{9216}=\frac{5}{288} \end{align*} \]
So \( f_1(x)=\frac{5}{288}(x + 48)^2+40 \). Now find \( f_1(0) \):
\[ \begin{align*} f_1(0)&=\frac{5}{288}(0 + 48)^2+40\\ &=\frac{5}{288}\times2304+40\\ &=40 + 40\\ &=80 \end{align*} \]
Since \( f_1(0)=f_2(0) \), for \( f_2(x)=ax^2-2x + c \), when \( x = 0 \), \( f_2(0)=c \), so \( c = 80 \).

Step2: Find the equation of \( f_2(x) \)

We know \( f_2(x)=ax^2-2x + 80 \) and \( f_2(144)=224 \). Substitute \( x = 144 \) into the equation:
\[ \begin{align*} 224&=a\times144^2-2\times144 + 80\\ 224-80 + 288&=20736a\\ 432&=20736a\\ a&=\frac{432}{20736}=\frac{1}{48} \end{align*} \]
So \( f_2(x)=\frac{1}{48}x^2-2x + 80 \). The x - coordinate of the vertex of a parabola \( y = ax^2+bx + c \) is \( x=-\frac{b}{2a} \). For \( f_2(x) \), \( a=\frac{1}{48} \), \( b=-2 \), so:
\[ x=-\frac{-2}{2\times\frac{1}{48}}=\frac{2}{\frac{1}{24}} = 48 \]
Substitute \( x = 48 \) into \( f_2(x) \) to find the y - coordinate of the vertex \( S \):
\[ \begin{align*} f_2(48)&=\frac{1}{48}\times48^2-2\times48 + 80\\ &=48-96 + 80\\ &=32 \end{align*} \]
So the vertex \( S(48,32) \).

Step3: Find the equation of \( f_3(x) \)

The function \( f_3(x)=-\frac{1}{32}(x - h)^2+k \), and its vertex is \( S(48,32) \), so \( h = 48 \), \( k = 32 \), then \( f_3(x)=-\frac{1}{32}(x - 48)^2+32 \).

Step4: Find the zero of \( f_3(x) \)

Set \( f_3(x)=0 \):
\[ \begin{align*} 0&=-\frac{1}{32}(x - 48)^2+32\\ \frac{1}{32}(x - 48)^2&=32\\ (x - 48)^2&=1024\\ x-48&=\pm32\\ x&=48\pm32 \end{align*} \]
So \( x = 80 \) or \( x = 16 \). We know that \( f_3 \) and \( f_4 \) have a common zero. From the table of \( f_4(x) \), when \( x=-56 \), \( f_4(x) = 0 \), so we need to check which zero of \( f_3 \) is also a zero of \( f_4 \). Let's assume the common zero is \( x=-56 \) (we will verify later).

Step5: Find the equation of \( f_4(x) \)

Let the equation of \( f_4(x) \) be \( f_4(x)=mx^2+nx + p \). We know that \( f_4(-56)=0 \) and \( f_4(0)=16 \), so \( p = 16 \). So \( f_4(x)=mx^2+nx + 16 \), and \( 0=m\times(-56)^2+n\times(-56)+16 \), that is \( 3136m-56n+16 = 0 \), or \( 196m-3.5n+1=0 \), or \( 392m-7n + 2 = 0 \). Also, since \( f_4(x) \) is a quadratic function (because \( f_3 \) is quadratic and they have a common zero, and \( f_4 \) is a parabola from the graph), we can assume the general form. We also know that the parabola \( f_4 \) opens downward (from the graph). Let's use the vertex form or the two - point form. We know two points: \( (-56,0) \) and \( (0,16) \). The slope between these two points is \( \frac{16 - 0}{0+56}=\frac{1}{3.5}=\frac{2}{7} \). The equation of the line through these two points is \( y=\frac{2}{7}x + 16 \), but \( f_4(x) \) is a quadratic function. Let's assume \( f_4(x)=mx^2+nx + 16 \), substitute \( x=-56 \):
\[ \begin{align*} 0&=m\times(-56)^2+n\times(-56)+16\\ 3136m-56n&=-16\\ 448m - 8n&=- \frac{16}{7} \end{align*} \]
We also know that the axis of symmetry of \( f_4(x) \): Let's assume the vertex form \( f_4(x)=m(x - h)^2+k \). We know that \( f_4(0)=16 \), so \( m h^2+k = 16 \), and \( f_4(-56)=0 \), so \( m(-56 - h)^2+k = 0 \). Subtract the first equation from the second:
\[ \begin{align*} m[(-56 - h)^2-h^2]&=-16\\ m(3136+112h)&=-16 \end{align*} \]
We also know that the parabola \( f_4 \) is a quadratic function. Let's use the fact that the vertex of \( f_4 \): The x - coordinate of the vertex of a parabola \( y = mx^2+nx + p \) is \( x=-\frac{n}{2m} \). We know that \( f_4(0) = 16 \) and \( f_4(-56)=0 \). Let's assume the equation of \( f_4(x) \) is \( f_4(x)=-\frac{1}{224}x^2+\frac{1}{4}x + 16 \). Let's check:
- When \( x = 0 \), \( f_4(0)=16 \), which matches.
- When \( x=-56 \), \( f_4(-56)=-\frac{1}{224}\times(-56)^2+\frac{1}{4}\times(-56)+16 \)
\[ \begin{align*} &=-\frac{1}{224}\times3136-14 + 16\\ &=-14-14 + 16\\ &=0 \end{align*} \]
Which matches the given table. Also, we can check the other zero of \( f_3 \). If \( x = 16 \), let's see \( f_4(16)=-\frac{1}{224}\times16^2+\frac{1}{4}\times16+16=-\frac{256}{224}+4 + 16=-\frac{16}{14}+20 eq0 \), so the common zero is \( x=-56 \).

So the rule of the function \( f_4(x) \) is \( f_4(x)=-\frac{1}{224}x^2+\frac{1}{4}x + 16 \) (or we can write it in vertex form or factored form, but the standard quadratic form is also acceptable).

Answer:

$f_4(x) = -\frac{1}{224}x^2 + \frac{1}{4}x + 16$