Question

Find the vertex of the graph of the function.

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.