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3 years ago

Write f(x)=8x^2-4x+11 in vertex form

Mathematics

1 answer:

Vlad [161]3 years ago

3 0

Answer:

$f(x)=8(x-\frac{1}{4})^{2}+\frac{21}{2}$

$f(x)=8(x-0.25)^{2}+10.5$

Step-by-step explanation:

we have

$f(x)=8x^{2}-4x+11$

This is a vertical parabola open upward (because the leading coefficient is positive)

The vertex is a minimum

Convert to vertex form

Factor the leading coefficient

$f(x)=8(x^{2}-\frac{1}{2}x)+11$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$f(x)=8(x^{2}-\frac{1}{2}x+\frac{1}{16})+11-\frac{1}{2}$

$f(x)=8(x^{2}-\frac{1}{2}x+\frac{1}{16})+\frac{21}{2}$

Rewrite as perfect squares

$f(x)=8(x-\frac{1}{4})^{2}+\frac{21}{2}$ ----> equation in vertex form

$f(x)=8(x-0.25)^{2}+10.5$

The vertex is the point (0.25,10.5)

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Which of the following points are more than 5 units from the point P(−2, −2)? Select all that apply. A A (2, 1) B B (4, −1) C C

netineya [11]

The distance between any 2 points P(a,b) and Q(c,d) in the coordinate plane, is given by the formula:<span>

<span> $|PQ|= \sqrt{ (a-c)^{2} + (b-d)^{2}}$ </span></span>

Using this formula we calculate the distances |PA|, |PB|, |PC|, |PD| and |PE| and compare to 5.

$|PA|= \sqrt{ (-2-2)^{2} + (-2-1)^{2}}= \sqrt{16+9}= \sqrt{25}=5$

$|PB|= \sqrt{ (-2-4)^{2} + (-2+1)^{2}}= \sqrt{36+1}= \sqrt{37} \approx 6$

$|PC|= \sqrt{ (-2-2)^{2} + (-2+3)^{2}}= \sqrt{16+1}= \sqrt{17}\approx4$

$|PD|= \sqrt{ (-2+6)^{2} + (-2+6)^{2}}= \sqrt{16+16}= \sqrt{32}\ \textgreater \ \sqrt{25}=5$

$|PE|= \sqrt{ (-2+5)^{2} + (-2-1)^{2}}= \sqrt{9+9}= \sqrt{16}=4$

Answer: B and D

3 0

3 years ago

elena55 [62]

Answer:

option 3, 45

Step-by-step explanation:

3x - 9° + 30° + 24° = 180° (angle sum property of a triangle)

3x - 9° + 54° = 180°

3x + 45° = 180°

3x = 180° - 45°

3x = 135°

x = 135/3

x = 45°

therefore, option 3 is the correct option

8 0

2 years ago

Which expression is equivalent to (r Superscript negative 7 Baseline) Superscript 6? r Superscript 42 StartFraction 1 Over r Sup

vodomira [7]

Power rule of exponents says that the power of the power of a exponent is equal to the multiplying both the powers. The expression which is equal to the given expression is,

$\dfrac{1}{(r)^{42}}$

The option B is the correct option.

Given information-

The given expression in the problem is,

$[(r)^{-7}]^5$

<h3>Power rule of exponents</h3>

Power rule of exponents says that the power of the power of a exponent is equal to the multiplying both the powers.

Suppose<em> </em><em>x</em> is a number with power <em>a. </em>The power is power of the power of number <em>x. </em>Then by the power rule of the exponents,

$(x^a)^b=x^{ab}$

Using the power rule of the exponents given expression can be written as,

$[(r)^{-7}]^6=(r)^{-7\times 6} 
$

$[(r)^{-7}]^6=(r)^{-42}$

<h3>Negative exponents rule</h3>

Negative exponents rule says that when a power of a number is negative, then write the number in the denominator with the same power with positive sign.Thus,

$[(r)^{-7}]^6=\dfrac{1}{(r)^{42}}$