Math  /  Algebra

QuestionQuestion 20
Let f(x)=2x2+7x+33x216x+16f(x)=\frac{2 x^{2}+7 x+3}{3 x^{2}-16 x+16} This function has: 1) A yy intercept at the point \square 2) xx intercepts at the point(s) \square 3) Vertical asymptotes at x=x= \square 4) Horizontal asymptote at y=y= \square Question Help: \square Video 11 \square Video 2 Check Answer

Studdy Solution

STEP 1

1. The function f(x)=2x2+7x+33x216x+16 f(x) = \frac{2x^2 + 7x + 3}{3x^2 - 16x + 16} is a rational function.
2. To find intercepts and asymptotes, we need to analyze both the numerator and the denominator.
3. The y y -intercept occurs when x=0 x = 0 .
4. The x x -intercepts occur where the numerator is zero and the denominator is not zero.
5. Vertical asymptotes occur where the denominator is zero and the numerator is not zero.
6. Horizontal asymptotes are determined by the degrees of the numerator and the denominator.

STEP 2

1. Find the y y -intercept.
2. Find the x x -intercepts.
3. Determine the vertical asymptotes.
4. Determine the horizontal asymptote.

STEP 3

To find the y y -intercept, set x=0 x = 0 in the function:
f(0)=2(0)2+7(0)+33(0)216(0)+16=316 f(0) = \frac{2(0)^2 + 7(0) + 3}{3(0)^2 - 16(0) + 16} = \frac{3}{16}
Thus, the y y -intercept is at the point (0,316) \left(0, \frac{3}{16}\right) .

STEP 4

To find the x x -intercepts, set the numerator equal to zero:
2x2+7x+3=0 2x^2 + 7x + 3 = 0
Use the quadratic formula x=b±b24ac2a x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} , where a=2 a = 2 , b=7 b = 7 , and c=3 c = 3 :
x=7±7242322 x = \frac{-7 \pm \sqrt{7^2 - 4 \cdot 2 \cdot 3}}{2 \cdot 2} x=7±49244 x = \frac{-7 \pm \sqrt{49 - 24}}{4} x=7±254 x = \frac{-7 \pm \sqrt{25}}{4} x=7±54 x = \frac{-7 \pm 5}{4}
This gives us two solutions:
x=7+54=24=12 x = \frac{-7 + 5}{4} = \frac{-2}{4} = -\frac{1}{2} x=754=124=3 x = \frac{-7 - 5}{4} = \frac{-12}{4} = -3
Thus, the x x -intercepts are at the points (12,0) \left(-\frac{1}{2}, 0\right) and (3,0) (-3, 0) .

STEP 5

To find the vertical asymptotes, set the denominator equal to zero:
3x216x+16=0 3x^2 - 16x + 16 = 0
Use the quadratic formula x=b±b24ac2a x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} , where a=3 a = 3 , b=16 b = -16 , and c=16 c = 16 :
x=16±(16)2431623 x = \frac{16 \pm \sqrt{(-16)^2 - 4 \cdot 3 \cdot 16}}{2 \cdot 3} x=16±2561926 x = \frac{16 \pm \sqrt{256 - 192}}{6} x=16±646 x = \frac{16 \pm \sqrt{64}}{6} x=16±86 x = \frac{16 \pm 8}{6}
This gives us two solutions:
x=16+86=246=4 x = \frac{16 + 8}{6} = \frac{24}{6} = 4 x=1686=86=43 x = \frac{16 - 8}{6} = \frac{8}{6} = \frac{4}{3}
Thus, the vertical asymptotes are at x=4 x = 4 and x=43 x = \frac{4}{3} .

STEP 6

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Both are degree 2, so the horizontal asymptote is the ratio of the leading coefficients:
y=23 y = \frac{2}{3}
The function has: 1) A y y -intercept at the point (0,316) \left(0, \frac{3}{16}\right) 2) x x -intercepts at the points (12,0) \left(-\frac{1}{2}, 0\right) and (3,0) (-3, 0) 3) Vertical asymptotes at x=4 x = 4 and x=43 x = \frac{4}{3} 4) A horizontal asymptote at y=23 y = \frac{2}{3}

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