How do you graph #y=x^2 + 6x + 8 #?

1 Answer
Aug 8, 2015

To draw the graph of the function, decide the range of x values in such a way, it includes the turning point of the curve.


To graph a function, we have to fix the range of values for x-variable.

First - Examine the function and have a rough idea about the shape of the curve.

It is a quadratic function. So it is a 'U' shaped curve and it has one turning point.

Second - Decide whether it is concave upwards or downwards.

Third -Find at what value of x the curve turns.

Showing the turning point is very important while we graph the function.

You know the general form of the quadratic function -
y = #ax^2# + bx + c

Since the co-efficient of #x^2# i.e., a is positive, the curve is concave upwards. In our case the coefficient of #x^2# is +1.

# -b/(2a)# gives the value of x- co-ordinate at which the curve turns. In our case it is # -6/(2xx1)# = -3.

Take three x-co-ordinate values on either side of '-3'.

In our case the possible x values are -6, -5, -4 , -3, -2, -1, 0.

Find the corresponding y values. These are the pair of points.
(-6, 8); (-5, 3); (-4, 0); (-3, -1); (-2, 0); (-1, 3); (0, 8)

Plot the pairs of points on a graph sheet and join all the points with the help of a smooth curve.
graph{y=x^2 + 6x + 8 [-10, 10, -5, 5]}