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lys-0071 [83]
3 years ago
14

Write the function in vertex form. identify the vertex.y = 2x^2 - 4x + 5​

Mathematics
1 answer:
jek_recluse [69]3 years ago
4 0

Answer:

\large\boxed{f(x)=2(x-1)^2+3}

Step-by-step explanation:

The vertex form of a quadratic function f(x) = ax² + bx + c:

f(x)=a(x-h)^2+k

Where (h, k) is a vertex.

h=\dfrac{-b}{2a},\ k=f(h)

We have:

f(x)=2x^2-4x+5\to a=2,\ b=-4,\ c=5

Calculate <em>h</em> and <em>k</em>:

h=\dfrac{-(-4)}{2(2)}=\dfrac{4}{4}=1\\\\k=f(1)\to k=2(1^2)-4(1)+5=2-4+5=3

Finally:

f(x)=2(x-1)^2+3

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D(x)=(x^2-12x+20)/(3x)
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Answers:

Vertical asymptote: x = 0

Horizontal asymptote: None

Slant asymptote: (1/3)x - 4

<u>Explanation:</u>

d(x) = \frac{x^{2}-12x+20}{3x}

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Discontinuities: (terms that cancel out from numerator and denominator):

Nothing cancels so there are NO discontinuities.

Vertical asymptote (denominator cannot equal zero):

3x ≠ 0  

<u>÷3</u>   <u>÷3 </u>

x ≠ 0

So asymptote is to be drawn at x = 0

Horizontal asymptote (evaluate degree of numerator and denominator):

degree of numerator (2) > degree of denominator (1)

so there is NO horizontal asymptote but slant (oblique) must be calculated.

Slant (Oblique) Asymptote (divide numerator by denominator):

  •        <u>(1/3)x - 4    </u>
  •    3x)    x² - 12x + 20
  •             <u>x²        </u>
  •                  -12x
  •                  <u>-12x         </u>
  •                             20 (stop! because there is no "x")

So, slant asymptote is to be drawn at (1/3)x - 4



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