Question

In the box, complete the first 4 steps for graphing the quadratic function given. (Use ^ on the keyboard to indicate an exponent.) Then print a sheet of graph paper and graph the quadratic function to turn in to your teacher. Be sure to label the axes and vertex.


y = -x^2 - 4x - 3

Answer 1

Why not begin by finding the coordinates of the vertex of this parabola? Rewrite the given quadratic equation in "vertex form," y=a(x-h)+k.

Next, find the vertical and horizontal asymptotes. Find the y-intercept. Share your work.

Answer 2

I don't know about h or k in the equation, I'm taught to use a b and c. So for this equation a= x b= 4 and c =3.

Answer 3

a=-x, rather.

Answer 4

Note that we generally follow a convention in algebra: letters such as a, b, c, etc., represent fixed numbers, whereas those such as x, y, z, etc., represent variables.

In y = -x^2 - 4x - 3, x and y are your variables, and a=-1, b=-4 and c=-3 are your constants.

Also frequently used is (h,k) to denote the coordinates of the vertex.

To get you started: from y = -x^2 - 4x - 3, please factor out the (-) sign in front of x^2 and 4x: -(x^2 + 4x). Then:

y = -x^2 - 4x - 3 = - (x^2 + 4x ) - 3

Your next move is to "complete the square." Familiar with that process?

Answer 5

I have been taught that yet.

Answer 6

If you must, you may choose a number of x-values and calculate the corresponding y-values. For example, with y = -x^2 - 4x - 3, if x=1, y=1(1)^2 -4(1) - 3 = -6.
Be certain to include x=0 in your list of x-values. At x=1, y=-6, so (1,-6) is a point on your graph. If you let x=0 and find y, you have then found the "y-intercept" of this quadratic. Find a few more points, plot them and then draw a nice, neat parabola through them.

Answer 7

Completing the square:
\[y=-x^2-4x-3=-(x^2+4x)-3=-(x^2+4x+(2)^2-(2)^2)-3\]
The middle term coefficient (the coefficient of the linear term) is positive, hence the factorized form must be in the form (a+b)^2
We also take out the (-2)^2, because the expansion of (a+b)^2 yields a b(constant) term that is strictly positive.

Hence,
\[y=-(x+2)^2+4-3\]
\[y=-(x+2)^2+1\]

for the equation
y=a(x-h)^2 + k,
(h,k) is the turning point / vertex.

Hence we read off our completed square form to obtain a turning point of (2,1).

Y-intercepts: Sub x=0 into the original function
x-intercepts: solve the equation for the roots i.e. the intersections with the x-axis

Answer 8

Additionally, the shape of a quadratic graph can be deduced from looking at the sign of the coefficient of the squared term.

Answer 9

For a negative coefficient of the squared term