Can someone help me answer both Task 1 and Task 2? I’m stuck and would really appreciate it.

Question

Task #1.

y=x^2-2x+15

What is the Vertex, y-intercept and x-intercepts?

Task #2.

An object is thrown upward from the top of an 80 foot building with an initial velocity of 64 feet per second. The equation that models the height of the object after x seconds is y= -16x^2 + 64x + 80

How much time will it

take the object to hit the ground after it is thrown from the top of the building?

Explain how you found the answer to #2. Show your work if you calculated the answer.

Answer 1

y=x^2-2x+15

complete the square

y=x^2-2x +15

take half of -2 and square it and add and subtract the answer

(-1)^2=1

y=x^2-2x+1 +15-1

y=(x-1)^2+14

vertex=(1,14)

y-intercept: set x=0 and solve for y

y=(0-1)^2+14

y=1+14

y=15 is the y-intercept

the vertex is (1,14) and the parabola opens upwards so there is no x-intercept

y=-16x^2+64x+80

16x^2-64x-80=y

when the object hits the ground the height is equal to 0

16x^2-64x-80=0

divide all terms on both sides by 16 to simplify the equation

x^2-4x-5=0

factor

(x-5)(x+1)=0

x-5=0

x=5 seconds (x=-1 also which you disregard because that would be -1 second which makes no sense)

Answer 2

Task #1:

For the vertex, the x coordinate is x = -b/2a = -(-2)/(2*1) = 1

To get the y coordinate plug in the x value found

y=(1)²-2(1)+15 = 14

Vertex: (1,14)

For the x intercepts, there are none, I saw this when I graphed to checked the vertex.

For the y intercept, plug in x=0 into the equation.

y=(0)²-2(0)+15 = 15

y intercept = (0,15)

Task #2:

To find when it hits the ground, set y=0 and solve.

y= -16x² + 64x + 80

0= -16x² + 64x + 80

To solve, lets try factoring. Factor out the GCF as a first step.

0 = -16(x² -4x - 5) Factor the remaining trinomial

0=-16(x-5)(x+1)

Set the factors equal to zero and solve

x-5=0 Add 5 to both sides

x-5+5=0+5

x=5

x+1=0 Subtract 1 from both sides

x+1-1=0-1

x=-1 Since time cannot be negative, this is not a valid solution

The object hits the ground 5 seconds after it was thrown off the building.

Answer 3

Task #1

For the y-intercept you need to plug in 0 for x and find the value for Y

( y=0^2-2(0)+15) - y intercept

For the x-intercept you need to plug in 0 for y and find the value for X

(0=x^2-2(x)+15) - x intercept

For vertex set the parabola to this ax^2+bx+c so a = 1, b=2 and c=15

Find the vertex of the parabola and store it in d => d=2a/b

Find the for e => c-(b^2/4a)

now substitute the value for the vertex formula a(x-h)^2 + e

The values h and e are your vertex (H,E);

Task #2

The height is going to be zero so you need to set the equation to zero

-16x^2+64x+80 = 0

Now here is makes sense to factor out this equation to:

-16(x^2-4x-5) =0

Here we can factor out the trinomial to:

-16(x+1)(x-5) = 0;

Then we can solve for both x's:

x+1=0 => x=-1

x-5=- => x=5

Since time can't be negative x=5 makes the most sense, and would be the time it takes.