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Bas_tet [7]
3 years ago
9

Find the vertex form of the following quadratic function. f(x)= x^2-11x+9

Mathematics
1 answer:
Shalnov [3]3 years ago
3 0

f(x) = (x - \frac{11}{2})² - \frac{85}{4}

the equation of a quadratic in vertex form is

y = a(x - h)² + k

where (h, k) are the coordinates of the vertex and a is a multiplier

given a quadratic in standard form : y = ax² + bx + c ( a ≠ 0 )

then the x- coordinate of the vertex is

x_{vertex} = - \frac{b}{2a}

f(x) = x² - 11x + 9 is in standard form

with a = 1, b = - 11 and c = 9

x_{vertex} = - \frac{-11}{2} = \frac{11}{2}

substitute this value into the equation for y- coordinate

y = (\frac{11}{2})² - 11(\frac{11}{2}) + 9

  = \frac{121}{4} - \frac{242}{4} + \frac{36}{4} = - \frac{85}{4}

f(x) = (x - \frac{11}{2})² - \frac{85}{4} ← in vertex form


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Answer:

4th choice \bold{f^{-1}(x) = \sqrt{x} + 4}

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<u>Definition of the inverse of a function</u>

A function g is the inverse of a function f if whenever y=f(x) then x=g(y). In other words, applying f and then g is the same thing as doing nothing. We can write this in terms of the composition of f and g as g(f(x))=x. The domain of f becomes the range of g and the range of f becomes the domain of g

To solve for the inverse of the function f(x) =\left(x-4\right)^2

Let y=\left(x-4\right)^2

\mathrm{Replace}\:x\:\mathrm{with}\:y \text{ and replace }\:y\:\mathrm{with}\:x

x=\left(y-4\right)^2

Switch sides
\left(y-4\right)^2=x

Take square roots on both sides
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Add 4 on both sides to solve for y

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To determine which one of these to be chosen not that in the given choices we can eliminate the first two since x cannot be negative

The third choice can also be eliminated since
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So the last answer choice is correct and the inverse off(x) = (x-4)^2

is given by  f^{-1}(x) = \sqrt{x} + 4

Answer:4th choice \bold{f^{-1}(x) = \sqrt{x} + 4}

Note

Domain of  (x-4)² is [4, ∞) since x ≥ 4 and (x-4)²  cannot be negative

Range of (x-4)²  is [0,  ∞)

Domain of \sqrt{x}\:+\:4 is [0, ∞)

Range of \sqrt{x}\:+\:4 is [4, ∞)

so indeed the domain of (x-4)² has become the range of \sqrt{x}\:+\:4 and the range of (x-4)² has become the domain of \sqrt{x}\:+\:4

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On a coordinate plane, a parabola opens up. It goes through (negative 8, negative 2), has a vertex at (negative 5, negative 6.5)
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Answer:

<h2>The top one</h2>

Step-by-step explanation:

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