Question

1.) Identify the axis of symmetry.
f(x) = x2 + 10x + 9

2.) Find the vertex for the parabola given below. f(x) = 3x2 + 12x + 16

3.) Find the quadratic function that includes each set of values.
(1, 0), (2, 5), (4, 21)

4.) f(x) = -2x2 - 3x + 35
Is this parabola concave up or concave down?

5.) A projectile is fired straight upward from the ground with an initial speed of 96 feet per second,then its height h after t seconds is given by the equation h(t) = -16t2 + 96t. Find the maximum height of the projectile.

Answer 1

#1
x^2+10x+9
= x^2+10x+25 + 8-25
= (x+5)^2 - 16
Since the vertex is at (-5,-16), the axis of symmetry is at x = -5

Or, you can consider that the roots are at -9 and -1, and the axis is midway between the roots, or at x = -5.

#2
3x^2+12x+16
= 3(x^2+4x) + 16
= 3(x^2+4x+4) + 16-3*4
= 3(x+2)^2 + 4
So, the vertex is at (-2,4)

#3
Plugging in values, we see that
a+b+c = 0
4a+2b+c = 5
16a+4b+c = 21
solve for a,b,c,and we see we have
x^2+2x-3

Or, since one root is at x=1,
y = a(x-1)(x-d)
5 = a(2-d)
21 = 3a(4-d)
a,d = 1,-3
y = (x-1)(x+3) = x^2+2x-3

#4 since the coefficient of x^2 is negative, the parabola opens downward

#5 since the vertex is at x = -b/2a, that will be when x = 3. h(3) = 144