Answer: The correct options are (1) (5,10), (2) (3,-3), (3) x = -1, (4)
, (5) 21s and (6) 0, -1, and 5.
Explanation:
Te standard form of the parabola is,
.....(1)
Where, (h,k) is the vertex of the parabola.
(1)
The given equation is,
![f(x)=(x-5)^2+10](https://tex.z-dn.net/?f=f%28x%29%3D%28x-5%29%5E2%2B10)
Comparing this equation with equation (1),we get,
and ![k=10](https://tex.z-dn.net/?f=k%3D10)
Therefore, the vertex of the graph is (5,10) and the fourth option is correct.
(2)
The given equation is,
![f(x)=3x^2-18x+24](https://tex.z-dn.net/?f=f%28x%29%3D3x%5E2-18x%2B24)
![f(x)=3(x^2-6x)+24](https://tex.z-dn.net/?f=f%28x%29%3D3%28x%5E2-6x%29%2B24)
To make perfect square add
, i.e.,
. Since there is factor 3 outside the parentheses, so subtract three times of 9.
![f(x)=3(x^2-6x+9)+24-3\times 9](https://tex.z-dn.net/?f=f%28x%29%3D3%28x%5E2-6x%2B9%29%2B24-3%5Ctimes%209)
![f(x)=3(x-3)^2-3](https://tex.z-dn.net/?f=f%28x%29%3D3%28x-3%29%5E2-3)
Comparing this equation with equation (1),we get,
and ![k=-3](https://tex.z-dn.net/?f=k%3D-3)
Therefore, the vertex of the graph is (3,-3) and the fourth option is correct.
(3)
The given equation is
![f(x)=4x^2+8x+7](https://tex.z-dn.net/?f=f%28x%29%3D4x%5E2%2B8x%2B7)
![f(x)=4(x^2+2x)+7](https://tex.z-dn.net/?f=f%28x%29%3D4%28x%5E2%2B2x%29%2B7)
To make perfect square add
, i.e.,
. Since there is factor 4 outside the parentheses, so subtract three times of 1.
![f(x)=4(x^2+2x+1)+7-4](https://tex.z-dn.net/?f=f%28x%29%3D4%28x%5E2%2B2x%2B1%29%2B7-4)
![f(x)=4(x+1)^2+3](https://tex.z-dn.net/?f=f%28x%29%3D4%28x%2B1%29%5E2%2B3)
Comparing this equation with equation (1),we get,
and ![k=3](https://tex.z-dn.net/?f=k%3D3)
The vertex of the equation is (-1,3) so the axis is x=-1 and the correct option is 2.
(4)
The given equation is,
![y=x^2+4x+7](https://tex.z-dn.net/?f=y%3Dx%5E2%2B4x%2B7)
To make perfect square add
, i.e.,
.
![f(x)=x^2+4x+4+7-4](https://tex.z-dn.net/?f=f%28x%29%3Dx%5E2%2B4x%2B4%2B7-4)
![f(x)=x^2+4x+4+7-4](https://tex.z-dn.net/?f=f%28x%29%3Dx%5E2%2B4x%2B4%2B7-4)
![f(x)=(x+2)^2+3](https://tex.z-dn.net/?f=f%28x%29%3D%28x%2B2%29%5E2%2B3)
Therefore, the correct option is 4.
(5)
The given equation is,
![h=-16t^2+672t](https://tex.z-dn.net/?f=h%3D-16t%5E2%2B672t)
It can be written as,
![h=-16(t^2-42t)](https://tex.z-dn.net/?f=h%3D-16%28t%5E2-42t%29)
It is a downward parabola. so the maximum height of the function is on its vertex.
The x coordinate of the vertex is,
![x=\frac{b}{2a}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7Bb%7D%7B2a%7D)
![x=\frac{42}{2}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B42%7D%7B2%7D)
![x=21](https://tex.z-dn.net/?f=x%3D21)
Therefore, after 21 seconds the projectile reach its maximum height and the correct option is first.
(6)
The given equation is,
![f(x)=3x^3-12x^2-15x](https://tex.z-dn.net/?f=f%28x%29%3D3x%5E3-12x%5E2-15x)
![f(x)=3x(x^2-4x-5)](https://tex.z-dn.net/?f=f%28x%29%3D3x%28x%5E2-4x-5%29)
Use factoring method to find the factors of the equation.
![f(x)=3x(x^2-5x+x-5)](https://tex.z-dn.net/?f=f%28x%29%3D3x%28x%5E2-5x%2Bx-5%29)
![f(x)=3x(x(x-5)+1(x-5))](https://tex.z-dn.net/?f=f%28x%29%3D3x%28x%28x-5%29%2B1%28x-5%29%29)
![f(x)=3x(x-5)(x+1)](https://tex.z-dn.net/?f=f%28x%29%3D3x%28x-5%29%28x%2B1%29)
Equate each factor equal to 0.
![x=0,-1,5](https://tex.z-dn.net/?f=x%3D0%2C-1%2C5)
Therefore, the zeros of the given equation is 0, -1, 5 and the correct option is 2.