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Question Find the range of the quadratic function f(x)=2(x−5)(x+5)f(x) = 2(x - 5)(x + 5).

Studdy Solution

STEP 1

Assumptions
1. The function given is f(x)=2(x−5)(x+5) f(x) = 2(x-5)(x+5) .
2. We are looking for the range of this function, which is the set of all possible output values (y-values) that the function can produce.
3. The function is a quadratic function in the form of f(x)=a(x−h)(x+k) f(x) = a(x-h)(x+k) , where a a , h h , and k k are constants.

STEP 2

Identify the form of the quadratic function.
The given function is in the form of f(x)=a(x−h)(x+k) f(x) = a(x-h)(x+k) , where a=2 a = 2 , h=5 h = 5 , and k=−5 k = -5 .

STEP 3

Determine the vertex of the parabola.
For a quadratic function in the form f(x)=a(x−h)2+k f(x) = a(x-h)^2 + k , the vertex is at the point (h,k) (h, k) . However, our function is not in this form. We need to rewrite it in the vertex form to find the vertex.

STEP 4

Rewrite the function in the vertex form.
To rewrite the function in the vertex form, we need to complete the square. However, in this case, the function is already factored, and we can find the vertex by looking at the symmetry of the roots.

STEP 5

Find the axis of symmetry.
The axis of symmetry for a quadratic function is at x=−b2a x = \frac{-b}{2a} when the function is in the standard form ax2+bx+c ax^2 + bx + c . However, since our function is factored, we can find the axis of symmetry by averaging the roots x=5 x = 5 and x=−5 x = -5 .
x=5+(−5)2=0x = \frac{5 + (-5)}{2} = 0

STEP 6

Find the y-coordinate of the vertex.
To find the y-coordinate of the vertex, we plug the x-coordinate of the axis of symmetry into the function.
f(0)=2(0−5)(0+5)f(0) = 2(0-5)(0+5)

STEP 7

Calculate the y-coordinate of the vertex.
f(0)=2(−5)(5)f(0) = 2(-5)(5)
f(0)=2(−25)f(0) = 2(-25)
f(0)=−50f(0) = -50

STEP 8

Determine the direction of the parabola.
Since the coefficient of the quadratic term a=2 a = 2 is positive, the parabola opens upwards.

STEP 9

Identify the range of the function.
Given that the parabola opens upwards and the vertex has a y-coordinate of -50, the range of the function is all real numbers greater than or equal to -50.

STEP 10

Write the range in interval notation.
The range of f(x)=2(x−5)(x+5) f(x) = 2(x-5)(x+5) is y≥−50 y \geq -50 , which can be written in interval notation as:
Range=[−50,∞)\text{Range} = [-50, \infty)
The range of f(x)=2(x−5)(x+5) f(x) = 2(x-5)(x+5) is [−50,∞) [-50, \infty) .

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