Question

graph the function $y = x^3 + 6x^2 + 2x - 11$. based on the graph, what is the largest possible $x$-value if $y = 0$? round your answer to the nearest hundredth, if necessary.

• 1.11
• 1.9
• 5.2
• 13.2

Explanation:

Step1: Analyze the function's end behavior

For the cubic function \( y = x^3+6x^2 + 2x-11 \), the leading term is \( x^3 \). As \( x
ightarrow+\infty \), \( y
ightarrow+\infty \) and as \( x
ightarrow-\infty \), \( y
ightarrow-\infty \). The function will cross the \( x \)-axis at least once. To find the roots, we can test the given options.

Step2: Test option 1.11

Substitute \( x = 1.11 \) into the function:
\( y=(1.11)^3+6\times(1.11)^2+2\times1.11 - 11 \)
\( y\approx1.3676+6\times1.2321 + 2.22-11 \)
\( y\approx1.3676 + 7.3926+2.22 - 11 \)
\( y\approx10.9802 - 11=- 0.0198\approx0 \)

Step3: Test option 1.9

Substitute \( x = 1.9 \) into the function:
\( y=(1.9)^3+6\times(1.9)^2+2\times1.9-11 \)
\( y = 6.859+6\times3.61+3.8 - 11 \)
\( y=6.859 + 21.66+3.8-11 \)
\( y=32.319 - 11 = 21.319
eq0 \)

Step4: Test option 5.2

Substitute \( x = 5.2 \) into the function:
\( y=(5.2)^3+6\times(5.2)^2+2\times5.2-11 \)
\( y=140.608+6\times27.04 + 10.4-11 \)
\( y=140.608+162.24+10.4 - 11 \)
\( y=313.248 - 11=302.248
eq0 \)

Step5: Test option 13.2

Substitute \( x = 13.2 \) into the function:
\( y=(13.2)^3+6\times(13.2)^2+2\times13.2-11 \)
\( y = 2299.968+6\times174.24+26.4 - 11 \)
\( y=2299.968+1045.44+26.4 - 11 \)
\( y=3371.808 - 11 = 3360.808
eq0 \)

Since when \( x = 1.11 \), the value of \( y \) is approximately 0 (very close to 0), and the other options do not give \( y\approx0 \) (except 1.11 which is very close), and we are looking for the root (where \( y = 0 \)), and among the options, 1.11 is the one that makes \( y\approx0 \). Also, since the function is cubic and we are looking for the largest possible \( x \)-value? Wait, no, wait. Wait, maybe I made a mistake. Wait, the cubic function \( y=x^3 + 6x^2+2x - 11 \). Let's find the derivative \( y'=3x^2 + 12x+2 \). The critical points are at \( x=\frac{-12\pm\sqrt{144 - 24}}{6}=\frac{-12\pm\sqrt{120}}{6}=\frac{-12\pm2\sqrt{30}}{6}=\frac{-6\pm\sqrt{30}}{3}\approx\frac{-6\pm5.477}{3} \). The critical points are approximately \( x\approx\frac{-6 + 5.477}{3}\approx - 0.174 \) and \( x\approx\frac{-6-5.477}{3}\approx - 3.826 \). So the function is increasing when \( x < - 3.826 \), decreasing between \( - 3.826 \) and \( - 0.174 \), and increasing when \( x > - 0.174 \). So the largest root will be the right - most root. When we test \( x = 1.11 \), \( y\approx0 \), \( x = 1.9 \) gives \( y>0 \), but wait, when \( x = 1.11 \), \( y\approx - 0.02 \), when \( x = 1.12 \):

\( y=(1.12)^3+6\times(1.12)^2+2\times1.12-11 \)
\( y\approx1.4049+6\times1.2544+2.24 - 11 \)
\( y\approx1.4049+7.5264+2.24 - 11 \)
\( y\approx11.1713 - 11 = 0.1713>0 \)

So the root is between \( 1.11 \) and \( 1.12 \), and among the given options, 1.11 is the closest to the root. The other options: 1.9, 5.2, 13.2 when substituted give \( y \) values far from 0. So the largest possible \( x \)-value (the right - most root) among the options is 1.11.

Answer:

A. 1.11