Question

How do you find the intercepts, vertex and the axis of symmetry?

Answer 1

x-intercepts: Where does the curve cross the x-axis?

Answer 2

so (0,-3) and (0,1)?

Answer 3

The x-coordinate goes first and then the y-coordinate. (x, y)

Answer 4

Ohh okat so (-3,0) and (1,0)?

Answer 5

yes.

Answer 6

Vertex of a parabola is either the lowest point on the graph or the highest point. In this case it is the lowest point. What are the coordinates of the lowermost point on the curve?

Answer 7

(-1,-4)? also how do you find the y-intercept?

Answer 8

Vertex at (-1, -4) is correct.

The y-intercept is where the curve crosses the y-axis.

Answer 9

(0,-3)?

Answer 10

Good!

Answer 11

Axis of symmetry here is the vertical line that passes through the vertex. This will split the curve exactly into two halves. What is the equation of a vertical line that passes through (-1,-4)?

Answer 12

-1?

Answer 13

The equation of a vertical line is x = constant. Here the constant is -1.
So x = -1 is the equation for the axis of symmetry.

Answer 14

ohh okay

Answer 15

Could you help me with one more?

Answer 16

sure.

Answer 17

I have to convert these between these three forms and Im not very good at it.

Answer 18

What is the starting equation or graph to convert? Does this question pertain to the previous graph?

Answer 19

No it doesn't I just have to convert the y=x^2-2x-3 to vertex form and intercept form and the same goes for the other columns

Answer 20

Just one small point about the previous question. Sometimes, when they ask for the x-intercepts it is sufficient to just say where it cuts the x-axis without putting it in the coordinate form. So we can say the x-intercepts are: -3 and 1 and the y-intercept is -3. Depends on your text book / teacher.

Answer 21

Ohh okay

Answer 22

y = x^2 - 2x - 3 is the standard form. To put it in vertex form, complete the square:
Add { -b/(2a) }^2 to both sides. -b/(2a) = -(-2) / (2 * 1) = 1, Square is also 1.
y + 1 = x^2 - 2x + 1 - 3
y + 1 = (x - 1)^2 - 3
y + 1 + 3 = (x - 1)^2
y + 4 = (x - 1)^2 is the vertex form.

Answer 23

To put y = x^2 - 2x - 3 in intercept form, factor the right hand side:
y = x^2 - 3x + x - 3
y = x(x - 3) + 1(x - 3)
y = (x+1)(x-3) is the intercept form. (right way you can tell the x-intercept for this equation will be -1 and -3).

Answer 24

To complete the square for y = ax^2 + bx + c
we need to add { -b / (2a) }^2 to both sides.

Answer 25

ohhh okay this is making more sense Im gonna see if I could do them next one on my own. Thank you(:

Answer 26

You are welcome.

Answer 27

Im having trouble getting the second one into the intercept form. so far I have\[y=-2x ^{2}-4x=3\]
\[y=-2(x ^{2}+2x+1)\]
y=-2(x+1)(x+1)

Answer 28

y - 5 = -2(x+1)^2 = -2(x^2 + 2x + 1) = -2x^2 - 4x - 2
y = -2x^2 - 4x + 3
This one does not factor nicely.

Answer 29

I thought if you took out the -2 it would?

Answer 30

Not really. The x-intercepts are irrational numbers and so it does not factor nicely.

Answer 31

brb

Answer 32

okay

Answer 33

If you want we can still factor it but it will look ugly with radicals and not so nice looking factors.

Answer 34

Can we because I need to know how to do this even if its not gonna look pretty.

Answer 35

We are going to use the identity: a^2 - b^2 = (a + b)(a - b).

Answer 36

y - 5 = -2(x + 1)^2
y = 5 - 2(x + 1)^2
2y = 10 - 4(x + 1)^2
2y = 10 - (2x + 2)^2
\(2y = (\sqrt{10})^2 - (2x + 2)^2\) (we have put it in the form a^2 - b^2
\(2y = (\sqrt{10}+2x+2)(\sqrt{10}-2x-2)\)
\(y = \frac 12(\sqrt{10}+2x+2)(\sqrt{10}-2x-2)\) is the intercept form.

Answer 37

Actually it is the factored form but not yet the intercept form.

Answer 38

\[
y = (\sqrt{10}/2 + x + 1) * 2(\sqrt{10}/2 - x - 1) \\
y = (x - (-1 - \sqrt{10}/2)) * (-2)(x + 1 - \sqrt{10}/2) \\
y = -2(x - (-1 - \sqrt{10}/2)) * (x - (-1 + \sqrt{10}/2)) \\
\]

Answer 39

wow that was a really ugly one thank you for all your help(:

Answer 40

You are welcome.