Question

Find the vertex of the graph of the function.

f(x) = (x - 5)2 + 10

(0, 5)
(10, 0)
(10, 5)
(5, 10)
Question 2(Multiple Choice Worth 1 points)
Find the vertex of the graph of the function.

f(x) = 3x2 - 18x + 24

(4, -2)
(-2, 4)
(-3, 3)
(3, -3)
Question 3(Multiple Choice Worth 1 points)
Find the axis of the graph of the function.

f(x) = 4x2 + 8x + 7

x = 3
x = -1
x = -2
x = 2
Question 4(Multiple Choice Worth 1 points)
Write the quadratic function in vertex form.

y = x2 + 4x + 7

y = (x - 2)2 + 3
y = (x - 2)2 - 3
y = (x + 2)2 - 3
y =(x + 2)2 + 3
Question 5(Multiple Choice Worth 1 points)
A projectile is thrown upward so that its distance above the ground after t seconds is h = -16t2 + 672t. After how many seconds does it reach its maximum height?

21 s
31.5 s
42 s
10 s
Question 6(Multiple Choice Worth 1 points)
Find the zeros of the function.

f(x) = 3x3 - 12x2 - 15x

1 and -5
0, -1, and 5
-1 and 5
0, 1, and -5

Answer 1

Answer: The correct options are  (1) (5,10), (2) (3,-3), (3) x = -1, (4) $y=(x+2)^2+3$, (5) 21s and (6) 0, -1, and 5.

Explanation:

Te standard form of the parabola is,

$f(x)=a(x-h)^2+k$        .....(1)

Where,  (h,k) is the vertex of the parabola.

(1)

The given equation is,

$f(x)=(x-5)^2+10$

Comparing this equation with equation (1),we get,

$h=5$ and $k=10$

Therefore, the vertex of the graph is (5,10) and the fourth option is correct.

(2)

The given equation is,

$f(x)=3x^2-18x+24$

$f(x)=3(x^2-6x)+24$

To make perfect square add $(\frac{b}{2a})^2$, i.e., $9$. Since there is factor 3 outside the parentheses, so subtract three times of 9.

$f(x)=3(x^2-6x+9)+24-3\times 9$

$f(x)=3(x-3)^2-3$

Comparing this equation with equation (1),we get,

$h=3$ and $k=-3$

Therefore, the vertex of the graph is (3,-3) and the fourth option is correct.

(3)

The given equation is

$f(x)=4x^2+8x+7$

$f(x)=4(x^2+2x)+7$

To make perfect square add $(\frac{b}{2a})^2$, i.e., $1$. Since there is factor 4 outside the parentheses, so subtract three times of 1.

$f(x)=4(x^2+2x+1)+7-4$

$f(x)=4(x+1)^2+3$

Comparing this equation with equation (1),we get,

$h=-1$ and $k=3$

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,

$y=x^2+4x+7$

To make perfect square add $(\frac{b}{2a})^2$, i.e., $2^2$.

$f(x)=x^2+4x+4+7-4$

$f(x)=x^2+4x+4+7-4$

$f(x)=(x+2)^2+3$

Therefore, the correct option is  4.

(5)

The given equation is,

$h=-16t^2+672t$

It can be written as,

$h=-16(t^2-42t)$

It is a downward parabola. so the maximum height of the function is on its vertex.

The x coordinate of the vertex is,

$x=\frac{b}{2a}$

$x=\frac{42}{2}$

$x=21$

Therefore,  after 21 seconds the projectile reach its maximum height and the correct option is first.

(6)

The given equation is,

$f(x)=3x^3-12x^2-15x$

$f(x)=3x(x^2-4x-5)$

Use factoring method to find the factors of the equation.

$f(x)=3x(x^2-5x+x-5)$

$f(x)=3x(x(x-5)+1(x-5))$

$f(x)=3x(x-5)(x+1)$

Equate each factor equal to 0.

$x=0,-1,5$

Therefore, the zeros of the given equation is 0, -1, 5 and the correct option is 2.