Question

Define the quadratic function f having x-intercepts 2 and 5, and y-intercept 5. So lost -_-

Answer 1

You want a quadratic function that has roots at x = 2 and x = 5, because the x-intercept is simply the point at which the value of y = 0.

I'll do a different set of roots, just so I'm not giving you the answer.

We can write a polynomial with roots at \(x = 1\) and \(x = -3\) as
\[P(x) = k(x-1)(x+3)\]because the roots are the place where the function = 0, and the product of such binomial terms will give us that behavior. The factor \(k\) is to allow us to scale the polynomial to hit the y intercept at x = 0. To find \(k\), we know that \[P(0) =k(0-1)(0+3) = 5\]\[k(-1)(3) = 5\]\[k = -\frac{5}{3}\]So for my set of roots, with a y-intercept of 5, the final polynomial would be \[P(x) = -\frac{5}{3}(x-1)(x+3) = -\frac{5}{3}(x^2+2x-3)\]
Not so hard, is it?

Answer 2

if you don't get that, ask me for an alternate way.

Answer 3

thanks guys! I'll give it a shot

Answer 4

If you don't mind, @hartnn, how about showing it to me, even if he does get it? :-)

Answer 5

okk..

Answer 6

let y= ax^2+bx+c
i'll also take different numbers.
Define the quadratic function f having x-intercepts 4 and 3, and y-intercept 2.

for y-intercept , put x=0
so, y = c = 2

for x-intercept, put y=0
ax^2 +bx+2 = 0
and since u have 2 x-intercepts, u get 2 equations in a and b .
a (4^2) +4b+2= 0
a (3^2) + 3b+2 =0
solve simultaneously to get a and b.

Answer 7

i know my method is not easier, its just an alternate way.

Answer 8

Oh, okay, sure. I did that with someone the other day to fit a parabola without even thinking about it, but of course it is equivalent. Which way would you prefer if you had say 5 or 6 roots?

Answer 9

5 to 6 roots means 4 to 5 variables, my method will become too cumbersome/complicated to solve.

Answer 10

Well, I'm always interested in different ways of thinking about problems, never know what is going to make the light bulb go on for any given student, thanks!