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2 years ago

(w-10x) (w+10x) Use the Identities to solve this question

Mathematics

1 answer:

Inessa [10]2 years ago

8 0

We are provided that –

(w-10x) (w+10x)

Applying identity –

a² - b² = (a + b)(a - b)

$\qquad$ $\twoheadrightarrow\bf(w-10x) (w+10x)$

$\qquad$ $\twoheadrightarrow\bf w^2 - (10x)^2$

$\qquad$ $\twoheadrightarrow\bf w^2 -100 x^2$

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Graph f(x) =-2x^+16x-30 by factoring to find the solutions, then find the coordinates of the vertex, and the axis of symmetry. I

fredd [130]

Answer:

Observe attached image

Function zeros:

(3, 0), (5, 0)

Vertex:

(4, 2)

Axis of symmetry:

 $x =4$ 

Step-by-step explanation:

First factorize the function

$f (x) = -2x ^ 2 + 16x-30$

Take -2 as a common factor.

$-2(x ^ 2 -8x +15)$

Now factor the expression $x ^ 2 -8x +15$

You must find two numbers that when you add them, obtain the result -8 and multiplying those numbers results in 15.

These numbers are -5 and -3

Then we can factor the expression in the following way:

$f (x) = -2(x-5)(x-3)$

The quadratic function cuts the x-axis at x = 3 and at x = 5.

Now we find the coordinates of the vertex.

For a function of the form $ax ^ 2 + bx + c$ the x coordinate of its vertex is:

$x = \frac{-b}{2a}$

In the function $f (x) = -2x ^ 2 + 16x-30$

$a = -2\\b = 16\\c = 30$

Then the vertice is:

$x = \frac{-16}{2(-2)}\\\\x = 4$

The y coordinate of the symmetry axis is

$y = f (4) = -2 (4) ^ 2 +16 (4) -30\\\\y = 2$

The axis of symmetry is a vertical line that cuts the parabola in two equal halves. This axis of symmetry always passes through the vertex.

Then the axis of symmetry is the line

$x = 4$

The solutions and the vertice written as ordered pairs are:

Function zeros:

(3, 0), (5, 0)

Vertex:

(4, 2)

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