How do you sketch a graph of this function? `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:
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.
Then, evaluate the function f(x)=x^2+8x+19 when x=h to get the value of k.
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.
Also, assign a value of x that is greater than h.
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```
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.