Find a quadratic function in standard form that has 3-1/4i as a zero and also passes through the point (1,-70).
To find a quadratic function in standard form that satisfies the given conditions, we can use the fact that complex conjugate zeros occur in pairs for quadratic functions with real coefficients.
Let's start by considering the given zero, 3 - 1/4i. Since complex conjugate zeros occur in pairs, the other zero must be its complex conjugate: 3 + 1/4i.
Now, we can use these zeros to construct our quadratic function. The standard form of a quadratic function is given by:
f(x) = a(x - r)(x - s)
where a is the leading coefficient, r and s are the zeros.
Substituting the given values:
f(x) = a(x - (3 - 1/4i))(x - (3 + 1/4i))
To simplify, we can multiply the factors together:
f(x) = a((x - 3 + 1/4i)(x - 3 - 1/4i))
Now, we can use the difference of squares formula, which states that (a + b)(a - b) = a^2 - b^2:
f(x) = a((x - 3)^2 - (1/4i)^2)
Expanding the squared term:
f(x) = a((x - 3)^2 - 1/16i^2)
Since i^2 = -1:
f(x) = a((x - 3)^2 + 1/16)
Now, we have a quadratic function in standard form. We can substitute the point (1, -70) into this equation to find the value of a, the leading coefficient.
-70 = a((1 - 3)^2 + 1/16)
Simplifying:
-70 = a((-2)^2 + 1/16)
-70 = a(4 + 1/16)
-70 = a((64 + 1)/16)
-70 = a(65/16)
To solve for a, we can multiply both sides by the reciprocal of (65/16), which is (16/65):
a = -70 * (16/65)
a = -112/13
Finally, substituting the value of a back into our quadratic function:
f(x) = (-112/13)((x - 3)^2 + 1/16)
Thus, the quadratic function in standard form with the given zero and passing through the point (1, -70) is:
f(x) = (-112/13)((x - 3)^2 + 1/16)