# How do you sketch a graph of this function? `f(x)=x^2+8x+19`

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`f(x)=x^2+8x+19`

Note that graph of a quadratic function is a parabola. To plot it, determine its vertex (h,k). To do so, apply the formula:

`h=-b/(2a)`

where a is the coefficient of x^2 and b coefficient of x.

In the function f(x)=x^2+8x+19, the values of a and b are 1 and 8, respectively. Plug-in these value to the formula to get h.

`h=-8/(2*1)=-8/2=-4`

Then, evaluate the function f(x)=x^2+8x+19 when x=h to get the value of k.

`k=f(h)=f(-4)=(-4)^2+8*(-4)+19`

`k=16-32+19=3`

Hence, the vertex of the parabola is (-4, 3).

Next, use additional two points to plot it. To do so, assign a value of x that is less than h. And solve for the corresponding value of y.

`x=-6` ,

`y=f(-6)=(-6)^2+8*(-6)+19=36-48+19=7`

Also, assign a value of x that is greater than h.

`x=-2` ,

`y=f(-2)=(-2)^2+8*(-2)+19=4-16+19=7`

So the other two points are (-6,7) and (-2,7).

Then, plot the vertex (-3,4) and the two points (-6,7) and (-2,7). Connect them and extend the parabola on both ends.

**Hence, the graph of the given function is:**

`1^(st)` step you have to know what it behaviour has:`f(x)=x^2+8x+19```

Then first step is the field of denifition, that, being a polynomial is all the Real field `R`

`2^(nd)` step is to find zeros in order to determinate where is positive or negative.

`x^2 + 8x + 19=0`

we can re-write so:

`x^2 +8x + 16 + 3= 0```

`(x+4)^2=-3`

This equation hasn't solution, so function is only o positive or negative.

since `f(0)=19` then the function is always positive.

Now le's study the slope to determine the behviour

`f'(x)= 2x +8`

the points we search are the points for `f'(x) =0`

We find, in this case only a point for:

`2x + 8=0` `x= -4`

Since the second derivative `f''(x)=2>0` for every x, this points out we've found a minimum point. the point `V(-4;3)`

Since it is of the second degree is a parable with vertex at the bottom at the point `V(-4;3)` .

as in the picture.