Question

[9.01] and [9.02] For y = x2 − 4x + 3,

Determine if the parabola opens up or down.
State if the vertex will be a maximum or minimum.
Find the vertex.
Find the x-intercepts.
Describe the graph of the equation.
Show all work and use complete sentences to receive full credit.

Answer 1

If there's no negative in front of the x^2 (provided y is not negative too) then it opens upwards
If the parabola opens upwards, then the vertex will be a minimum
To find the vertex, put your equation in graphing form by completing the square (tell me if that step confuses you)
You'll get y=(x-2)^2-1 which is y=a(x-h)^2+k, and the vertex is at (h,k)
So this one is at (2,-1)
To find the x intercepts, set the y=0 and then factor the x's, you'll get
0=(x-3)(x-1) and set each of those =0, you'll get x=1,3 and those are your x intercepts
Describe the graph... Opens upwards, is normal width, drops below the x axis 2 and moves left 1

Answer 2

thank you so much, can you help me with another one?

Answer 3

[9.06] Fill in the missing term so that the quadratic equation has a graph that opens up, with a vertex of
(– 3, – 25), and x intercepts at x = -8 and x = 2. (Do not include the negative sign in your answer.)

y = x2 + 6x − ___

Numerical Answers Expected!

Answer 4

Since it gives you the x intercepts, you can think back to how you would have found the x intercepts
You would have gotten two things factored that when you set them equal to 0 you got -8 and 2, so what would those two factored things be? (x+8) and (x-2)
Now if you work back even more, and multiply those two back together to unfactor, you'll get the original equation, which would be x^2+6x-16