Help with quadratic function

[hellomath2552]

Find a quadratic function in vertex form that has -1/2-4i as a zero and y intercept (0,-10)

Please help

 

[mmm4444bot]

How may we help you? What have you thought about so far? Where are you stuck?

 

[hellomath2552]

well i got to ax^2+x+16 1/4
not sure if thats correct

 

[mmm4444bot]

That's close. How did you do it?

Complex roots with an imaginary part always come in conjugate pairs. Hence, we know that -1/2+4i is also a root.

Factors take the form (x-root); the polynomial factors as:

a(x-root1)(x-root2)

Substituting the roots and expanding:

a(x^2 + x + 65/4)

Note that we don't do arithmetic with mixed numbers; leave 65/4 as an improper fraction.

The given y-intercept allows you to find a.

Complete the square, and write the vertex form.

 

[pappus]

Find a quadratic function in vertex form that has -1/2-4i as a zero and y intercept (0,-10)

Please help

You are looking for the equation of a parabola in vertex form which is:

\(\displaystyle \displaystyle{y = a(x-h)^2+k}\)

with the vertex V(h, k)

1. You know that the x-coordinate of the vertex is the mean of the two zeros: \(\displaystyle \displaystyle{x_V=-\frac12}\)

The equation reads now:

\(\displaystyle \displaystyle{y = a \left(x+\frac12 \right)^2+k}\)

2. Using the values of the zero and the y-intercept you'll get two equations:

\(\displaystyle \displaystyle{0=a(-4i)^2+k}\)

\(\displaystyle \displaystyle{-10=a \left(\frac12 \right)^2+k}\)

Solve this system of simultaneous equations for (a, k)