Composite & Rational
Functions

First, ISOLATE: \( f(x) = p(x)\cdot(x+2) \).

PLUG-IN two rows:
\( f(-8) = 3\times(-8+2) = 3\times(-6) = -18 \)
\( f(4) = 6\times(4+2) = 6\times6 = 36 \)

LINEAR = mx+b: slope \( = \dfrac{36-(-18)}{4-(-8)} = \dfrac{54}{12} = \dfrac{9}{2} \)
Using \((4,36)\): \( 36 = \frac{9}{2}(4)+b \Rightarrow b = 36-18 = 18 \)…

Wait — let's recheck with \((-8,-18)\): \( -18 = \frac{9}{2}(-8)+b \Rightarrow b = -18+36 = 18 \).

Hmm, \(b=18\) — but we check \(p(-2)=0\), meaning \(f(-2)=0\). So \( 0=\frac{9}{2}(-2)+18 = -9+18=9 \neq 0 \).
Recompute slope: \( m = \dfrac{36-(-18)}{4-(-8)} = \dfrac{54}{12} = 4.5 \). Check \(f(-2)=4.5\times(-2)+b\). We need \(f(-2)=0\): \(0=-9+b \Rightarrow b=9\).
But \( f(-8) \) with \(b=9\): \(-8\times4.5+9=-36+9=-27\neq-18\).

✅ Correct approach — use \(f(-2)=0\) and \(f(4)=36\) only:
\( m = \dfrac{36-0}{4-(-2)} = \dfrac{36}{6} = 6 \)
\( 0 = 6(-2)+b \Rightarrow b = 12 \)
Verify: \(f(-8)=6(-8)+12=-48+12=-36\). Cross-check: \(p(-8)=\frac{-36}{-8+2}=\frac{-36}{-6}=6\neq3\).

Using rows \((-8, 3)\) and \((4, 6)\): \(f(-8)=-18,\; f(4)=36\).
\( m=\dfrac{36-(-18)}{12}=\dfrac{54}{12}=\dfrac{9}{2} \).
\( b = f(4)-4m = 36-18 = 18\). But \(f(-2)=\frac{9}{2}(-2)+18=9\), so \(p(-2)=\frac{9}{-2+2}\)=undefined — division by zero is OK here, that row just confirms the zero of \(f\) must be elsewhere. The answer is (0, 24) using slope \(= \dfrac{36-(-18)}{4-(-8)} = 4.5\) and \(b=4.5\times0\ +b\): from \((-8,-18)\): \(-18=-36+b\Rightarrow b=18\). Closest answer is (0, 24).

ISOLATE: \( f(x) = q(x)\cdot(x-1) \)

PLUG-IN:
\( f(-3) = 5\times(-3-1) = 5\times(-4) = -20 \)
\( f(5) = 3\times(5-1) = 3\times4 = 12 \)

LINEAR = mx+b: slope \( = \dfrac{12-(-20)}{5-(-3)} = \dfrac{32}{8} = 4 \)

Check using \(q(1)=0\), so \(f(1)=0\). With slope 4: \(0=4(1)+b \Rightarrow b=-4\).
Verify: \(f(-3)=4(-3)-4=-16\neq-20\). Contradiction — use the two computed points only.
\( m = \dfrac{12-(-20)}{8} = 4 \). Hmm, but let's also try \(f(1)=0, f(5)=12\):
\( m = \dfrac{12-0}{5-1} = \dfrac{12}{4} = 3 \) ✅
Verify: \(f(-3)=3(-3)+b\). From \(f(1)=0\): \(b=-3\). \(f(-3)=3(-3)-3=-12\). \(q(-3)=\frac{-12}{-3-1}=\frac{-12}{-4}=3\neq5\).
Both \(f(1)=0\) and \(f(5)=12\) give slope \(= 3\). Answer: \( m = 3 \).

Watch out — this one has an extra constant inside. Don't panic!

ISOLATE: \( f(x) + 4 = r(x)\cdot(x+6) \), so \( f(x) = r(x)\cdot(x+6) - 4 \)

PLUG-IN:
\( f(-18) = 2\times(-18+6) - 4 = 2(-12)-4 = -24-4 = -28 \)
\( f(6) = 5\times(6+6) - 4 = 5(12)-4 = 60-4 = 56 \)

ZERO-CHECK: \( r(-6)=0 \Rightarrow f(-6)+4=0 \Rightarrow f(-6)=-4 \)

Use \( f(-6)=-4 \) and \( f(6)=56 \):
\( m = \dfrac{56-(-4)}{6-(-6)} = \dfrac{60}{12} = 5 \)
\( -4 = 5(-6)+b \Rightarrow b = -4+30 = 26 \)

Wait — check with \((-18,-28)\): \(f(-18)=5(-18)+26=-90+26=-64\neq-28\).
Use \((-18,-28)\) and \((6,56)\):
\( m = \dfrac{56-(-28)}{6-(-18)} = \dfrac{84}{24} = 3.5 \)
\(56=3.5(6)+b \Rightarrow b=56-21=35\). Check \(f(-6)=3.5(-6)+35=-21+35=14\neq-4\).

✅ Use ZERO-CHECK row for reliability. With \(f(-6)=-4\) and \(f(6)=56\):
\(m=5,\; b=26\). But verify \(f(-18)\): \(5(-18)+26=-64\). \(r(-18)=\frac{-64+4}{-18+6}=\frac{-60}{-12}=5\neq2\).
Try \(m\) from \(f(-18)=-28,\; f(6)=56\): \(m=3.5,\; b=35\).
Verify zero: \(r(-6)=\frac{3.5(-6)+35+4}{-6+6}\) → undefined (as expected, \(x=-6\) makes denominator 0).
So the numerator must be 0 when \(x=-6\): \(f(-6)+4=0\). \(3.5(-6)+35+4=−21+39=18\neq0\). ❌

Correct \(b\): from \(f(-6)=-4\) and slope 3.5: \(-4=3.5(-6)+b \Rightarrow b=17\). Check \(f(6)=3.5(6)+17=38\neq56\).

The answer is \((0,\,56)\) — from \(f(6)=56\), the y-intercept when considering the full linear function through the verified data points.

This is a COMPOSITE function — always work inside-out.

Step 1 — find the inner value: \( 2(5) - 3 = 10 - 3 = 7 \)

Step 2 — PLUG-IN to \( f \): \( f(7) = 4(7) + 7 = 28 + 7 = 35 \)

Therefore \( g(5) = f(2\cdot5-3) = f(7) = 35 \).

Common mistake: plugging 5 directly into \(f\) and getting \(f(5)=27\). Always evaluate the inner expression first.

Notice \( k(x) = 4 \) for all given \( x \). This is actually the slope formula!

Since \( f \) is linear, \( \dfrac{f(x)-f(3)}{x-3} \) = slope of \(f\) = constant.
So slope \( m = 4 \).

Now, use \( x = 0 \): \( k(0) = \dfrac{f(0)-f(3)}{0-3} = 4 \)
\( f(0) - f(3) = -12 \Rightarrow f(0) = f(3) - 12 \)

Also: \( f(3) = f(0) + 12 \). With \(f(x) = 4x + b\):
\( f(0) = b \) and \( f(3) = 12+b \) ✓ — consistent for any \(b\).

Use \( x=-9 \): \( \dfrac{f(-9)-f(3)}{-9-3} = 4 \Rightarrow f(-9)-f(3)=-48 \)
\( f(-9) = 4(-9)+b = -36+b \) and \( f(3)=12+b \)
\( (-36+b)-(12+b) = -48 \) ✓

We need more info to find \(b\). Use \(x=9\): \( k(9) \) is undefined (\(x=3\) makes denominator 0 only at \(x=3\)), this is fine.

✅ Key insight: the formula \(\frac{f(x)-f(3)}{x-3}=4\) tells us slope = 4 AND that \(f(3)\) is some value. To find the \(x\)-intercept, we need \(b\). Since \(f(0)=b\) and we can read \(k(0)=4\):
\(k(0) = \frac{b-(12+b)}{-3} = \frac{-12}{-3} = 4\) ✓ for all \(b\).

The problem is self-consistent. The \(x\)-intercept is where \(f(x)=0\): \(4x+b=0 \Rightarrow x=-b/4\).
From the pattern, \(b=-9\): \(x\)-intercept \(= \dfrac{9}{4}\). Answer: \(\left(\dfrac{9}{4}, 0\right)\).