The cost (in millions of dollars) for a 30 -second ad during the TV broadcast of a major sporting event can be approximated by $$ 0.488 x^{2}-4.678 x+26.6 $$ $$ x+4 $$ If this trend continues, will the cost of an ad reach $$ \$ 9 $$ million by 2028 ?

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We interpret the cost function as
$$
C(x)=\frac{0.488x^2-4.678x+26.6}{x+4},
$$
where $$ x $$ is the number of years after a specified base year. (It is common in such problems to let $$ x=0 $$ correspond to a particular starting year; here we assume that $$ x=28 $$ corresponds to 2028.) We wish to determine when the cost reaches $$ \$9 $$ million, that is, when
$$
\frac{0.488x^2-4.678x+26.6}{x+4}=9.
$$

**Step 1. Set Up the Equation**

To find the value of $$ x $$ for which $$ C(x)=9 $$, we solve:
$$
\frac{0.488x^2-4.678x+26.6}{x+4} = 9.
$$

**Step 2. Clear the Denominator**

Multiply both sides by $$ x+4 $$ (we assume $$ x\neq -4 $$):
$$
0.488x^2 - 4.678x + 26.6 = 9(x+4).
$$
Expand the right side:
$$
0.488x^2 - 4.678x + 26.6 = 9x + 36.
$$

**Step 3. Rearrange the Equation**

Bring all terms to one side:
$$
0.488x^2 - 4.678x - 9x + 26.6 - 36 = 0.
$$
Combine like terms:
$$
0.488x^2 - 13.678x - 9.4 = 0.
$$

**Step 4. Solve the Quadratic Equation**

We use the quadratic formula $$ x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} $$ with
$$
a=0.488,\quad b=-13.678, \quad c=-9.4.
$$
First, compute the discriminant:
$$
\Delta = b^2-4ac = (-13.678)^2 - 4(0.488)(-9.4).
$$
Calculate:
$$
(-13.678)^2 \approx 187.2,
$$
and
$$
-4(0.488)(-9.4)= 4(0.488)(9.4) \approx 18.3488.
$$
Thus,
$$
\Delta\approx 187.2+18.35\approx205.55.
$$
Taking the square root:
$$
\sqrt{\Delta}\approx \sqrt{205.55}\approx 14.34.
$$
Now, apply the quadratic formula:
$$
x=\frac{13.678\pm14.34}{2(0.488)}.
$$
We obtain two solutions:
$$
x_1=\frac{13.678+14.34}{0.976}\approx\frac{28.018}{0.976}\approx28.7,
$$
$$
x_2=\frac{13.678-14.34}{0.976}\approx\frac{-0.662}{0.976}\approx-0.678.
$$
Since $$ x $$ represents a number of years after the base year, we discard the negative solution. Thus, the cost reaches $$ \$9 $$ million when
$$
x\approx28.7.
$$

**Step 5. Relate $$ x $$ to the Year 2028**

If we assume (as is common in such problems) that $$ x=28 $$ corresponds exactly to 2028, then the cost reaches $$ \$9 $$ million when $$ x\approx28.7 $$, which is slightly later than 2028 (about 70% of a year after the 28‑year mark). In other words, by 2028 the cost would be slightly less than $$ \$9 $$ million.

To illustrate, we can evaluate the cost function at $$ x=28 $$:
$$
C(28)=\frac{0.488(28)^2-4.678(28)+26.6}{28+4}.
$$
Compute the numerator:
$$
28^2=784,\quad 0.488\times784\approx382.592,
$$
$$
4.678\times28\approx130.984,
$$
$$
382.592-130.984+26.6\approx278.208.
$$
Then,
$$
C(28)\approx\frac{278.208}{32}\approx8.694\text{ million}.
$$
This confirms that in 2028 the cost is about $$ \$8.69 $$ million, which is below $$ \$9 $$ million.

**Conclusion**

If the trend continues according to the given model, the cost of a 30‑second ad will not have reached $$ \$9 $$ million by 2028 but rather slightly after that year.
By 2028, the cost of a 30-second ad will not reach \$9 million; it will be approximately \$8.69 million.

The cost (in millions of dollars) for a 30 -second ad during the TV broadcast of a major sporting event can be approximated by \( 0.488 x^{2}-4.678 x+26.6 \) \( x+4 \) If this trend continues, will the cost of an ad reach \( \$ 9 \) million by 2028 ?

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