Question

The function f(x) = x2 + 10x – 3 written in vertex form is f(x) = (x + 5)2 – 28. What are the coordinates of the vertex?

Answer 1

Answer:

Option A is correct

(–5, –28)

Step-by-step explanation:

For a quadratic equation:

$y=ax^2+bx+c$

We know that convert of quadratic into vertex form, using the method of completing the square is,

$y = a(x-h)^2+k$

where,

(h, k) is the vertex.

As per the statement:

The function $f(x) = x^2 + 10x -3$

written in vertex form is:

$f(x) = (x+5)^2-28$

From the above definition we have;

h = -5 and k = -28

Therefore,  the coordinates of the vertex is, (-5, -28)

Answer 2

The coordinates of the vertex is $\boxed{\bf (-5,-28)}$.

Further explanation:

Given:

The function $f(x)$ is as follows:

$\boxed{f(x)=x^{2}+10x-3}$  

Vertex form of the function $f(x)$ is as follows:

$\boxed{f(x)=(x+5)^{2}-28}$

Definition of vertex form:

The vertex form of a quadratic function is as follows:

$\boxed{f(x)=a(x-h)^{2}+k}$    …… (1)

Here, $(h,k)$ is the vertex of the parabola.

The standard form of the quadratic equation is as follows:

$\boxed{f(x)=ax^{2}+bx+c}$

If the function $f(x)$ is written in vertex form then $(h,k)$ is the vertex of the parabola and $x=h$ is the axis of symmetry.

Calculation:

To convert the function $f(x)$ from standard form to vertex form first convert the function $f(x)$ into the complete square form as shown below,

$\begin{aligned}f(x)&=x^{2}+10x-3\\&=x^{2}+(2\cdot x\cdot 5)-3+5^{2}-5^{2}\\&=x^{2}+(2\cdot x\cdot 5)+5^{2}-3-25\\&=(x+5)^{2}-28\end{aligned}$

Label the above equation as follows:

$\boxed{f(x)=(x+(-5))^{2}-28}$     ......(2)

Compare equation (1) and equation (2) to obtain the value of $h$ and   $k$.

The value of $h$ and $k$ are as follows:

$\boxed{\begin{aligned}h&=-5\\k&=-28\end{aligned}}$

Therefore, the vertex of the parabola is at the point $(-5,-28)$.

A direct formula to find the vertex $(h,k)$ of the parabola is,

$\boxed{h=-\dfrac{b}{2a}\text{ and }k=f(h)}$

The horizontal shift $h$ is calculated as follows:

$\begin{aligned}h&=-\dfrac{10}{2\cdot 1}\\&=\dfrac{-10}{2}\\&=-5\end{aligned}$

 

The vertical shift $k$ is calculated as follows:

$\begin{aligned}k&=f(-5)\\&=(-5)^{2}+(10\cdot (-5)-3)\\&=25-50-3\\&=-25-3\\&=-28\end{aligned}$

Therefore, the coordinates of the vertex is at the point $(-5,-28)$.

Thus, the coordinates of the vertex is $\boxed{\bf (-5,-28)}$.

Learn more:

1. Learn more about vertex form brainly.com/question/1286775

2. Learn more about quadratic function brainly.com/question/1332667

Answer details:

Grade: High school

Subject: Mathematics

Chapter: Quadratic function

Keywords: Function, vertex form, quadratic function, coordinates, x2+10x-3, f(x) standard form, parabola, vertical shift, horizontal shift, degree 2, highest power is 2, quadratic equation.