Question

Identify the​ vertex, the axis of​ symmetry, the maximum or minimum​ value, and the range of the parabola for y=x^2+6x+13

Answer 1

Answer:

Step-by-step explanation:

First we can determine the x value of our vertex via the equation:

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

Note that in general a quadratic equation is such that:

$ax^2+bx+c$

In this case a,b and c are the coefficients and so a=1, b=6 and c=13.

Therefore we can determine the x component of the vertex by plugging in the values known and so:

$x=\frac{-b}{2a}=\frac{-6}{2(1)}=\frac{-6}{2}=-3$

Now we can determine the y-component of our vertex by plugging in the x-component to the equation and so:

$f(x)=x^2+6x+13\\\\f(-3)=(-3)^2+6(-3)+13\\\\f(-3)=4$

Therefore our vertex is (-3,4).  Now in vertex our x component determines is the axis of symmetry so the equation for axis of symmetry is:

x=-3

Similarly, the y-component of our vertex is the minimum or maximum.  In this case it is the minimum you can determine this because a is positive meaning that the parabola will point up, and so the equation for the  minimum is:

y=4

The range of the formula is the smallest y-value meaning the minimum y=4 and all real numbers that are more than 4, mathematically:

Range = All real numbers greater than or equal to 4.