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

Question

Abigail Allen · Accepted Answer

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]}

Genesis Allen · Answer

undefined To graph the equation y = x^2 + 6x + 8:

1. Identify the vertex using the formula: (-b/2a, f(-b/2a)).
2. Calculate the y-intercept by substituting x = 0 into the equation.
3. Find the x-intercepts by solving the equation for y = 0.
4. Plot the vertex, y-intercept, and x-intercepts on the coordinate plane.
5. Determine the direction of the parabola (opening upwards or downwards) based on the coefficient of x^2.
6. Sketch the parabola passing through the plotted points, maintaining symmetry.
7. Optionally, plot additional points if needed for accuracy.