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2 years ago

Which form of a quadratic function is easiest to use when finding the maximum or minimum value of the function?

Mathematics

2 answers:

BaLLatris [955]2 years ago

7 0

Answer:

Vertex form

Step-by-step explanation:

There are several forms of the quadratic equation

Standard form: y = ax^2 + bx + c which is useful for the quadratic equation and the axis of symmetry

Factored form: y = (ax - c)(bx - d) which will give us the zeros

and

Vertex form: y = a(x - h)2 + k where ( h,k) is the vertex

The maximum and minimum would be the value of k

It would be maximum when a >0 and minimum when a<0

alexandr402 [8]2 years ago

4 0

Answer:

ax² + bx + c

Step-by-step explanation:

The form of a quadratic equation that is easy to use when finding the maximum or minimum value of the function is ax² + bx + c.

Suppose a quadratic function:

f(x) = 2x² - 8x + 9

Use ( -b/2a , f(-b/2a) ).

-b/2a

a = 2

b = -8

-(-8)/2(2)

8/4

= 2

f(2) = 2(2)² - 8(2) + 9

f(2) = 2(4) - 8(2) + 9

f(2) = 8 - 16 + 9

f(2) = 1

The minimum value of this quadratic function is (2, 1).

It represents a minimum value because a > 0.

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Suppose that 30% of all houses need a paint job. Also, 15% of all houses need both a paint job and a new roof. Further, 7% of al

ivann1987 [24]

Answer:

0.0105

Step-by-step explanation:

given that 30% of all houses need a paint job. Also, 15% of all houses need both a paint job and a new roof. Further, 7% of all houses that need a new roof also need new windows.

If total houses are 100, then 30 require a paint along. 15 require both paint and new roof.

Out of 15 houses which require new roof 7% need new windows

i.e. no of houses which need new windows = $15*7%\\=1.05$

If there are total 100 houses, 1.05 houses require a new window

So probability that a randomly selectede house needs a new window

= 0.0105

8 0

2 years ago

Pls remember to give your answer in metres

spin [16.1K]

Answer: 500m

Explanation:
If the scale of the drawing is 1:500 then the real playground should be 500:1 which is 500m

5 0

2 years ago

On a coordinate plane, a circle has a center at (0, 0). Point (3, 0) lies on the circle.

9966 [12]

Answer:

The correct option is;

No, the distance from (0, 0) to (2, √6) is not 3 units

Step-by-step explanation:

The given parameters of the question are as follows;

Circle center (h, k) = (0, 0)

Point on the circle (x, y) = (3, 0)

We are required to verify whether point (2, √6) lie on the circle

We note that the radius of the circle is given by the equation of the circle as follows;

$Distance \, formula = \sqrt{\left (x_{2}-x_{1} \right )^{2} + \left (y_{2}-y_{1} \right )^{2}}$

Distance² = (x - h)² + (y - k)² = r² which gives;

(3 - 0)² + (0 - 0)² = 3²

Hence r² = 3² and r = 3 units

We check the distance of the point (2, √6) from the center of the circle (0, 0) as follows;

$\sqrt{\left (x_{2}-x_{1} \right )^{2} + \left (y_{2}-y_{1} \right )^{2}} = Distance$

Therefore;

(2 - 0)² + (√6 - 0)² = 2² + √6² = 4 + 6 = 10 = √10²

$\sqrt{\left (2-0 \right )^{2} + \left (\sqrt{6} -0 \right )^{2}} = 10$

Which gives the distance of the point (2, √6) from the center of the circle (0, 0) = √10

Hence the distance from the circle center (0, 0) to (2, √6) is not √10 which s more than 3 units hence the point (2, √6), does not lie on the circle.

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