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viktelen [127]
3 years ago
12

What is the vertex of the graph of y = 5(x + 4)2 + 3?

Mathematics
2 answers:
pychu [463]3 years ago
6 0
The vertex form: y = a(x - h)² + k

(h; k) - coordinates of the vertex

Therefore:

y = 5(x + 4)² + 3
vertex: (-4; 3)

Answer: A: -4, 3
vovikov84 [41]3 years ago
4 0
A: -4,3 The x coordinate of the vertex is given by x = -b/2a and the y coordinate of the vertex is given by y= f( -b/2a), we need to transform the equation to the form ax^2 + bx + c We have: 5(x+4)^2+3 = 5x^2+ 20x + 23 Then we substitute x= -20/2*5 y= f(-20/2*5) = -4 = f(-4) =3 So V(-4,3)
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Find the third order maclaurin polynomial. Use it to estimate the value of sqrt1.3
vodka [1.7K]

\sqrt{1+3x}=1+\frac{3}{2} x-\frac{9}{8} x^{2} + \frac{81}{8}x^{3} is the maclaurin polynomial and estimate value of \sqrt{1.3} is 1.14. This can be obtained by using the formula to find the maclaurin polynomial.

<h3>Find the third order maclaurin polynomial:</h3>

Given the polynomial,

f(x)=\sqrt{1+3x}=(1+3x)^{\frac{1}{2} }

The formula to find the maclaurin polynomial,

f(0)+\frac{f'(0)}{1!}x+\frac{f''(0)}{2!}x^{2} + \frac{f'''(0)}{3!}x^{3}

Next we have to find f'(x), f''(x) and f'''(x),

  • f'(x) = \frac{3}{2}(1+3x)^{-\frac{1}{2} }
  • f''(x) =-\frac{9}{4}(1+3x)^{-\frac{3}{2} }
  • f'''(x) = \frac{81}{8}(1+3x)^{-\frac{5}{2} }

By putting x = 0 , we get,

  • f(0)=(1+3(0))^{\frac{1}{2} }=1
  • f'(0) = \frac{3}{2}(1+3(0))^{-\frac{1}{2} }=\frac{3}{2}
  • f''(0) =-\frac{9}{4}(1+3(0))^{-\frac{3}{2} }=-\frac{9}{4}
  • f'''(0) = \frac{81}{8}(1+3(0))^{-\frac{5}{2} }=\frac{81}{8}

Therefore the maclaurin polynomial by using the formula will be,

\sqrt{1+3x}=f(0)+\frac{f'(0)}{1!}x+\frac{f''(0)}{2!}x^{2} + \frac{f'''(0)}{3!}x^{3}

\sqrt{1+3x}=1+\frac{3}{2} x-\frac{9}{8} x^{2} + \frac{81}{8}x^{3}

To find the value of \sqrt{1.3}  we can use the maclaurin polynomial,

\sqrt{1.3} is  \sqrt{1+3x} with x = 1/10,

\sqrt{1+3(1/10)}=1+\frac{3}{2} (1/10)-\frac{9}{8} (1/10)^{2} + \frac{81}{8}(1/10)^{3}

\sqrt{1+3(1/10)}=\frac{18247}{16000} = 1.14

Hence \sqrt{1+3x}=1+\frac{3}{2} x-\frac{9}{8} x^{2} + \frac{81}{8}x^{3} is the maclaurin polynomial and estimate value of \sqrt{1.3} is 1.14.

Learn more about maclaurin polynomial here:

brainly.com/question/24188694

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