1. Let f(x) = -2x^{2} - 12x + 5. Find

(a) the x- and y-intercepts

x-intercepts:

The x-intercepts are the solutions of 0 = -2x^{2} - 12x + 5. Since the

factorization is not immediately apparent, we will use the quadratic

formula to find the roots.

So the x-intercepts are
and x.

y-interceps:

y = -2(0)^{2} - 12(0) + 5 = 5

So the y-intercept is y = 5.

(b) the axis of symmetry.

(c) the vertex.

We know that the vertex is contained on axis of symmetry x = -3.

So the vertex is (-3, f(-3)) = (-3, 23).

(d) Is the vertex a max or a min?

We know that a = -2, so the parabola is downward opening and the

vertex is a maximum.

2. Find the equation y = ax^{2} + bx + c of the quadratic
function with vertex

(-2, 3) and passing through (2,-1).

We know that any quadratic function can be expressed in the form f(x) =

a(x-h)^{2}+k where (h, k) is the vertex of the parabola. Using the fact that

we want the vertex to be at (-2, 3), we therefore get that:

y = a(x + 2)^{2} + 3:

So all that remains is to find the value of a. To do this, we use the fact

that (2,-1) is on the graph of f.

So putting this value of a into the function, we get:

3. Setting up equations.

(a) If the area of a rectangle is 30cm, express the perimeter as a function

of width.

We want an equation for the perimeter P. So we start by
writing the

function in terms of two variables:

P = 2w + 2L

We also know that the area of the rectangle is 30cm, so we have the

equation 30 = wL. We now solve this equation in terms of L to get

and substitute this back into the function P. This gives

b) A rectangle is inscribed in a circle of radius 10.
Express the perimeter

as a function of width.

We want an equation for the perimeter P. So we start by
writing the

function in terms of two variables:

P = 2w + 2L.

We now need to find a relationship between the variables w and L,

and the only other information we have is that the radius of the circle

is 8. To understand how we will get the relationship, I slightly change

our picture.

We now see more clearly that the diameter, w and L are the
sides

of a right triangle. Using the Pythagorean Theorem, we see that

20^{2} = w^{2} + L^{2} . We now solve this equation in terms f L to get L =

(Why didn't I choose
?) and substitute

this ak into the function P. This gives

(c) A right triangle has vertices at the origin and at the
points (x, 0)

and (0, y) (with x and y positive). If the hypotenuse of the triangle

passes through the point (3, 5), express the area of the triangle as a

function of x.

We want an equation for the area A. So we start by writing
the

function in terms of two variables:

We now need to find a relationship between x and y. We note that

the points (0, y), (3, 5) and (x, 0) must be collinear , so the slope

between (0, y) and (3, 5) must be the same as the slope between (3, 5)

and (x, 0). Thus we get:

When we substitute this back into our function, we get that

(d) The sum of two numbers is 25. Find the value of the
two numbers

that maximizes their product.

i. Express the product P of the two numbers as a function of one

variable.

We start with

P = xy

where x and y are the numbers. We also know that x + y = 25.

If we solve this equation for y, we get that y = 25-x. If we now

substitute this back into the function P, we get:

P = x(25 - x)

P = 25x - x^{2
}

ii. This is a quadratic function. Find the maximum of P.

We know that P = 25x - x^{2} is a quadratic with a = -1, so we

know the graph is a downward opening parabola. Therefore, the

maximum of the function will be at the vertex of the parabola.

We know the x-coord of the vertex is

We know from (i) that y = 25 - x, so we know that the two

numbers that maximize P are and
.

The maximum value of the product is .