Question

Find the x intercepts of the parabola with vertex (1,-17) and y intercept (0,-16)

Answer 1

please help me I will fan and medal

Answer 2

http://www.mathwarehouse.com/geometry/pa...
Vertex Form of Equation
y= a(x-h)^2 + k

Knowing (h,k) is the vertex, and the vertex is (1,-17)

y = (x-1)^2 - 17
y = x^2 - 2x + 1 - 17
y = x^2 - 2x - 16

Knowing: y = ax 2 + bx + c
a = 1
b = -2
c = -16

Solving for x intercepts: http://www.analyzemath.com/quadratics/ve...
x1 = [ - b + sqrt (b^2 - 4 a c) ] / 2 a
x2 = [ - b - sqrt (b^2 - 4 a c) ] / 2 a

Solving:
x1 = [2 + sqrt(4 + 64)] / 2
x1 = (2 + sqrt(68)) / 2

x2 = [ - b - sqrt (b^2 - 4 a c) ] / 2 a
x2 = [2 - sqrt(4 + 64)] / 2
x2 = (2 - sqrt(68)) / 2


Therefore, rounding to the nearest hundredth: We know y is zero because they are x-intercepts.
(5.12, 0), (-3.12, 0)

Answer 3

so all of that is rounded to the nearest hundredth?

Answer 4

@SpooderWoman

Answer 5

yep

Answer 6

sorry it took so long to reply I had to take my dogs out :p

Answer 7

that's okay :) can you help me do the same thing with vertex -1,-16 and y intercept 0,-15?

Answer 8

Sure!

Answer 9

Okay: first, write the equation of the parabola in the required form:
(y - k) = a·(x - h)²

Here, (h, k) is given as (-1, -16).
So you have:
(y + 16) = a · (x + 1)²

Unfortunately, a is not given. However, you do know one additional point on the parabola: (0, -15):

-15 + 16 = a· (0 + 1)²
.·. a = 1

.·. the equation of the parabola in vertex form is
y + 16 = (x + 1)²

The x-intercepts are the values of x that make y = 0. So, let y = 0:

0 + 16 = (x + 1)²
16 = (x + 1)²

We are trying to solve for x, so take the square root of both sides - but be CAREFUL!

± 4 = x + 1 ...... remember both the positive and negative roots of 16......

Solving for x:
x = -1 + 4, x = -1 - 4
x = 3, x = -5.

Or, if you prefer, (3, 0), (-5, 0).

Answer 10

Are you taking FLVS?

Answer 11

FLVS???

Answer 12

Online class

Answer 13

These seem like the questions I had on my virtual school :p

Answer 14

Yes I'm taking summer courses with Ed options Academy trough Plato

Answer 15

Nice

Answer 16

Any more questions you need help with?

Answer 17

Hold on for a sec?

Answer 18

Yep :)

Answer 19

vertex 1,245 and y intercept 0,240

Answer 20

y = a(x-1)² + 245
240 = a+245
a = -5

y = -5(x-1)² + 245 = 0
-5(x² - 2x + 1) + 245 = 0
x² - 2x - 48 = 0
(x-8)(x+6) = 0
x = -6, 8

x-intercepts: (-6,0) and (8,0)

Answer 21

Want me to explain that one more?

Answer 22

no I think I got that one

Answer 23

Alright :) (Thanks for the testimony)

Answer 24

no problem :) vertex (1,1) y intercept (0,-3)

Answer 25

vertex form of equation for a parabola: y=A(x-h)^2+k, (h,k)=coordinates of vertex
solving for A:
-3=A(0-1)^2+1
A=-4
equation: y=-4(x-1)^2+1

x-intercepts
set y=0
-4(x-1)^2+1=0
(x-1)^2=1/4
x-1=±√(1/4)=±1/2
x=1±1/2
x-intercepts:
x=1/2
x=3/2

Answer 26

so how w\[(x _{1},y _{1}),(x _{2},y _{2})\]ould you write that as

Answer 27

For which one

Answer 28

for the one you just did

Answer 29

oh I think it would be (3/2,0) (0,-3) To show the x-intercept and the y-intercept.

Answer 30

okay thank you that was my last one

Answer 31

No problem! Good Luck with your schooling!