F(x) =5x^2-20x+3 how to find minimum

Mathematics

1 answer:

Leya [2.2K]3 years ago

3 0

Answer:

(2,-17) should be the minimum.

Step-by-step explanation:

The minimum of a quadratic function occurs at $x=-\frac{b}{2a}$ . If a is positive, the minimum value of the function is $f(-\frac{b}{2a})$

$f_{min}x=ax^2+bx+c$ occurs at $x=-\frac{b}{2a}$

Find the value of $x=-\frac{b}{2a}$

x = 2

evaluate f(2).

replace the variable x with 2 in the expression.

$f(2)=5(2)^2-20(2)+3$

simplify the result.

$f(2)=5(4)-20(2)+3$

$f(2)=20-40+3$

$f(2)=-17$

The final answer is -17

Use the x and y values to find where the minimum occurs.

HOPE THIS HELPS!

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The main cable of a suspension bridge forms a parabola, described by the equation y = a(x - h)2 + k, where y is the height in fe

nirvana33 [79]

Refer to the diagram shown below.

When x = 30 ft, the cable is at 15 ft, therefore y(30) = 15.
That is,
a(30 - h)² + k = 15 (1)

Also, because the distance between the supports is 90 ft, therefore
y(0) = 6 ft, and y(90) = 6 ft
That is,
a(-h)² + k = 6 (2)
a(90 - h)² + k = 6 (3)

From (2) and (3), obtain
a(90 - h)² = ah²
90² - 180h + h² = h²
180h = 90²
h = 45 ft.

From (1) and (2), obtain
225a + k = 15
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Therefore
1800a = -9
a = - 0.005
k = 15 - 225(-0.005) = 16.125 ft

Answer:
The equation for the cable is
y = - 0.005(x - 45)² + 16.125

A graph of the solution verifies that the solution is correct.