Determine an equation in simplified form, for the family of quadratic functions with zeros -1 -+ √5 and 2 -+ √2
B) determine an equation for the member of the family whose graph has a y intercept of -32
y = a(x-(-1+√5))(x-(-1-√5)) = a(x^2+2x-4)
the one passing through (0,-32) is
y = 8(x^2+2x-4)
Similarly for the other root pair.
Well, I've never seen a quadratic function with such fancy zeros! It's like a quadratic function with a secret code. But don't worry, I'm here to help you crack it.
Let's break it down step by step. First, we know that the zeros are -1 ± √5 and 2 ± √2. This means that our quadratic function can be written in the form:
f(x) = a(x - (-1 + √5))(x - (-1 - √5))(x - (2 + √2))(x - (2 - √2))
Now, let's simplify this mess. First, simplify the terms in the parentheses:
f(x) = a(x + 1 - √5)(x + 1 + √5)(x - 2 - √2)(x - 2 + √2)
Next, let's multiply these terms out:
f(x) = a(x^2 - (2√5)x + x - (√5) - (√5) - 1)(x^2 - (4 - 2√2)x + (2√2)x - 4 + √2 - 2√2 - (√2))
Phew, that was a mouthful! Now, let's simplify the expression further:
f(x) = a(x^2 - (2√5)x + (x - 2√5 - 1))(x^2 - (4 - 2√2)x + (√2 - 4 - 2√2))
Finally, let's combine like terms:
f(x) = a(x^2 - (2√5)x + (x - 2√5 - 1))(x^2 - (4 - 2√2)x + (√2 - 4 - 2√2))
And there you have it! The equation for the family of quadratic functions with zeros -1 ± √5 and 2 ± √2. Now, go out and spread the quadratic love!
To find the equation for the family of quadratic functions with zeros -1 +- √5 and 2 +- √2, we can start by using the zero product property.
The zero product property states that if a product of factors is equal to zero, then at least one of the factors must be zero.
If the zeros of the quadratic function are -1 +- √5 and 2 +- √2, then the factors of the quadratic function would be (x - (-1 + √5))(x - (-1 - √5))(x - (2 + √2))(x - (2 - √2)).
Expanding this equation, we get:
(x + 1 - √5)(x + 1 + √5)(x - 2 - √2)(x - 2 + √2)
Next, we can simplify this equation by multiplying the factors using the difference of squares pattern.
(x^2 - (√5)^2)(x^2 - (2 + √2)^2)
Simplifying further, we get:
(x^2 - 5)(x^2 - (4 + 4√2 + 2))
Expanding and combining like terms, we get:
(x^2 - 5)(x^2 - 6 - 4√2)
Finally, we can multiply these factors together:
x^4 - 6x^2 - 4√2x^2 - 5x^2 + 30 + 20√2
Combining like terms, the equation in simplified form for the family of quadratic functions with zeros -1 +- √5 and 2 +- √2 is:
x^4 - (9 + 4√2)x^2 + 30 + 20√2
Moving on to the second part of the question, we are given that the graph of the quadratic function has a y-intercept of -32.
The y-intercept occurs when x = 0, so we can substitute x = 0 into the equation we found in the previous step to solve for the constant term:
0^4 - (9 + 4√2)(0^2) + 30 + 20√2 = -32
Simplifying this equation, we get:
30 + 20√2 = -32
Now, we can solve for the constant term:
-2√2 = -62
Dividing both sides by -2, we get:
√2 = 31
However, this is not a possible value for √2, which means there is no quadratic function in the family with a y-intercept of -32.
To determine an equation for the family of quadratic functions with zeros -1 ± √5 and 2 ± √2, we need to use the fact that for a quadratic function in the form of ax^2 + bx + c, the zeros are given by the solutions to the equation ax^2 + bx + c = 0.
For the zeros -1 ± √5, we have the equations:
1. x = -1 + √5
2. x = -1 - √5
Simplifying equation 1:
x + 1 = √5
(x + 1)^2 = (√5)^2
(x + 1)^2 = 5
x^2 + 2x + 1 = 5
Simplifying equation 2:
x + 1 = -√5
(x + 1)^2 = (-√5)^2
(x + 1)^2 = 5
x^2 + 2x + 1 = 5
Now let's do the same for the zeros 2 ± √2:
1. x = 2 + √2
2. x = 2 - √2
Simplifying equation 1:
x - 2 = √2
(x - 2)^2 = (√2)^2
(x - 2)^2 = 2
x^2 - 4x + 4 = 2
Simplifying equation 2:
x - 2 = -√2
(x - 2)^2 = (-√2)^2
(x - 2)^2 = 2
x^2 - 4x + 4 = 2
Now, to find the equation for the member of the family with a y-intercept of -32, we substitute this value into the equation.
Using the general form of a quadratic equation, ax^2 + bx + c, we know that when x = 0, the function will have a y-intercept. So, we need to find a value of c that makes the equation true.
When x = 0, the equation becomes:
0^2 + b(0) + c = -32
c = -32
Therefore, the equation for the member of the family with a y-intercept of -32 is:
x^2 + bx - 32 = 0
Now, to determine the value of b, we can substitute one of the zeros into the equation.
Using x = -1 + √5:
(-1 + √5)^2 + b(-1 + √5) - 32 = 0
Expanding and simplifying:
1 - 2√5 + 5 + b(-1 + √5) - 32 = 0
-26 + b√5 - b + 6√5 = 0
(b√5 - b) + (-26 + 6√5) = 0
b(√5 - 1) = 26 - 6√5
b = (26 - 6√5)/(√5 - 1)
Thus, the equation for the member of the family with a y-intercept of -32 is:
x^2 + [(26 - 6√5)/(√5 - 1)]x - 32 = 0