Question

f(x)= x² +5x-2 Find the vertex and the quadratic function in standard form

Answer 1

The given function is

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

The vertex has two coordinates, h, and k. To find h we use the following formula.

$h=-\frac{b}{2a}$

Where a = 1 and b = 5.

$h=-\frac{5}{2(1)}=-\frac{5}{2}$

Then, we evaluate the function when x = -5/2 to find k.

$\begin{gathered} k=f(-\frac{5}{2})=(-\frac{5}{2})^2+5(-\frac{5}{2})-2 \\ k=\frac{25}{4}-\frac{25}{2}-2 \end{gathered}$

Let's find the least common factor between the denominators.

4 2 | 2

2 1 | 2

1

The least common factor would be 2*2 = 4, let's use it.

$\begin{gathered} k=\frac{25-2\cdot25-4\cdot2}{4} \\ k=\frac{25-50-8}{4} \\ k=-\frac{33}{4} \end{gathered}$

Once we know the value of h and k, we can deduct the vertex.

Therefore, the vertex is (-5/2, -33/4).

The standard form of a quadratic function is

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

In this case, a = 1, h = -5/2, and k = -33/4.

So, the standard form of f(x) is

$f(x)=(x+\frac{5}{2})^2-\frac{33}{4}$