Hey guys my brother told me you guys are very helpful so help me

I'm having trouble on this

Take the last two digits in your id and add them together. Call this number n.

a)
Write an equation for a quadratic function that has x-intercepts (n,0) and (n+1,0)
and a y-intercept (0,n+2). Provide the formula and the graph.

b) Write an equation for a quadratic function that has vertex (n,-n) and y-intercept (0,3).
Provide the formula and the graph.

N = 7

how would I even start this equation?

Thx

2. Originally Posted by Afterme
Hey guys my brother told me you guys are very helpful so help me

I'm having trouble on this

Take the last two digits in your id and add them together. Call this number n.

a)
Write an equation for a quadratic function that has x-intercepts (n,0) and (n+1,0)
and a y-intercept (0,n+2). Provide the formula and the graph.

b) Write an equation for a quadratic function that has vertex (n,-n) and y-intercept (0,3).
Provide the formula and the graph.

N = 7

how would I even start this equation?

Thx
Say the last two digits of your ID are 3 and 9. (I just picked them randomly.) So 3 + 9 = 12.

So
a) We need a quadratic function passing through the points (12, 0), (13, 0), and (0, 14).

The general quadratic function (assuming we want the parabola to open up or down) is
$y = ax^2 + bx + c$

So we know that
$0 = a(12)^2 + b(12) + c \implies 144a + 12b + c = 0$
and
$0 = a(13)^2 + b(13) + c \implies 169a + 13b + c = 0$
and
$14 = a(0)^2 + b(0) + c \implies c = 14$

Thus
$144a + 12b + 14 = 0$
and
$169a + 13b + 14 = 0$

Solve for a and b.

b) The vertex form of a quadratic would be more handy here:
$y = a(x - h)^2 + k$
This is a parabola with a vertex at (h, k).

So we need a vertex at (12, -12) and y-intercept at (0, 3). From the vertex:
$y = a(x - 12)^2 - 12$

Putting in the point (0, 3):
$3 = a(0 - 12)^2 - 12 = 144a - 12$
which you can solve for a.

-Dan