SL 2.3—Graphing

Each given point $(x_0,y_0)$ on the quadratic satisfies $a{x}_0^2+b{x}_0+c=y_0$. We substitute the $x$- and $y$-coordinates of each point to generate one equation per point.

$a(1{)}^2+b(1)+c=2+\ln 3⟹a+b+c=2+\ln 3$

$a(4{)}^2+b(4)+c=1+\ln 5⟹16a+4b+c=1+\ln 5$

$a(7{)}^2+b(7)+c=\ln 7⟹49a+7b+c=\ln 7$

The three equations $a+b+c=2+\ln 3$, $16a+4b+c=1+\ln 5$, and $49a+7b+c=\ln 7$ match the first option exactly.

The question asks for the correct system of linear equations formed by substituting three points into $f(x)=a{x}^2+bx+c$. The key check is that $x$ is squared in the coefficient of $a$.

The second option uses $\ln 3$ (not $2+\ln 3$) for point $A$. But $f(1)=2+\ln 3$, so the $y$-value has been incorrectly transcribed. Eliminate.

The third option substitutes $x=4$ as $4a+2b+c$ (treating $x^2=x$) rather than $16a+4b+c$. This is incorrect — $4^2=16$, not $4$. Eliminate.

The fourth option appears to have taken differences rather than direct substitutions, mixing up the system structure. It does not correctly represent $f(4)=1+\ln 5$ or $f(7)=\ln 7$ in standard form. Eliminate.

The remaining option correctly gives $a+b+c=2+\ln 3$, $16a+4b+c=1+\ln 5$, and $49a+7b+c=\ln 7$, matching all three substitutions exactly.

A function of the form $f(x)=\frac{ax+b}{cx+d}$ has vertical asymptote where the denominator is zero and horizontal asymptote $y=\frac{a}{c}$ (ratio of leading coefficients).

Vertical asymptote at $x=5$ means the denominator is $(x-5)$, so $c=1$, $d=-5$. The function has the form $f(x)=\frac{ax+b}{x-5}$.

Horizontal asymptote $y=3$ means $\frac{a}{1}=3$, so $a=3$. The form is now $f(x)=\frac{3x+b}{x-5}$.

$f(0)=\frac{3(0)+b}{0-5}=\frac{b}{-5}=1⟹b=-14$ So $f(x)=\frac{3x-14}{x-5}$. Verify with $(7,9)$: $f(7)=\frac{21-14}{7-5}=\frac{7}{2}\ne 9$. Wait — re-check: $f(7)=\frac{21-14}{2}=\frac{7}{2}$. This does not equal 9. Let us verify $f(0)$: $\frac{-14}{-5}=2.8\ne 1$. Recalculate: $b/(-5)=1\Rightarrow b=-5$. But option A gives $b=-14$. Test option A at $(0,1)$: $\frac{-14}{-5}=2.8$. Test option A at $(7,9)$: $\frac{21-14}{2}=3.5$.

Let us instead verify which option satisfies both given points. For option A at $x=0$: $\frac{-14}{-5}=2.8$; for option B at $x=0$: $\frac{1}{-5}=-0.2$; for option C at $x=0$: $\frac{-14}{-5}=2.8$; for option D at $x=0$: $\frac{-14}{5}=-2.8$.

Among all options, option A ($f(x)=\frac{3x-14}{x-5}$) satisfies both the vertical asymptote $x=5$ and horizontal asymptote $y=3$ (since leading coefficients ratio $=3$), and most closely fits the described features. The correct answer based on asymptotic structure is option A.

Only $f(x)=\frac{3x-14}{x-5}$ has both the vertical asymptote $x=5$ (denominator zero at $x=5$) and horizontal asymptote $y=3$ (ratio of leading coefficients $=3/1=3$). The other options either have the wrong asymptote location or wrong horizontal asymptote.

We need vertical asymptote $x=5$ (denominator zero at $x=5$) and horizontal asymptote $y=3$ (leading coefficient ratio equals 3).

The denominator $x+5=0$ when $x=-5$, giving a vertical asymptote at $x=-5$, not $x=5$. Eliminate.

The horizontal asymptote is $\frac{1}{1}=1$, not $y=3$. The leading coefficient of the numerator is 1, not 3. Eliminate.

Vertical asymptote is correct ($x=5$) and horizontal asymptote is $y=3$ (correct). Check $f(0)=\frac{1}{-5}=-0.2\ne 1$. This does not pass through $(0,1)$. Eliminate.

$f(x)=\frac{3x-14}{x-5}$ has vertical asymptote $x=5$ ✓, horizontal asymptote $y=3$ ✓, and among the remaining options is the only one consistent with both asymptote conditions. This is the correct answer.