Graph the parabola
x^2+8x-5y+21=0
I can't graph this for you.
What do you not understand about graphing?
Pick values for x and solve for y.
After you get a handful of x,y values, plot the points for your graph.
x^2 + 8x ..... + 21 = 5y
x^2 + 8x + 16 - 16 + 21 = 5y
(x+4)^2 + 5 = 5y
y = (1/5)(x+4)^2 + 1
Can you read the vertex from that?
Can you pick some suitable values of x to find a few other points for graphing ?
don't I need to put it in y= formula first?
Sorry didn't see Reiny's post
I'm not sure I will look over my notes to see what I come up with.
To graph the parabola, we need to rearrange the equation into the standard form of a parabola, which is y = ax^2 + bx + c.
Let's start by isolating the y term:
x^2 + 8x - 5y + 21 = 0
Rearranging the equation, we get:
-5y = -x^2 - 8x - 21
Now, divide both sides of the equation by -5 to solve for y:
y = (1/5)x^2 + (8/5)x + (21/5)
Now we have the equation in the standard form, where a = 1/5, b = 8/5, and c = 21/5.
To graph the parabola, we can use a graphing calculator or plotting points manually.
If we choose to plot points manually, we can select various x-values, substitute them into the equation to find the corresponding y-values, and then plot the points on a graph.
Let's choose a range of x-values, such as -10 to 10, and calculate the corresponding y-values:
When x = -10,
y = (1/5)(-10)^2 + (8/5)(-10) + (21/5)
y = 4 + (-16) + 21/5
y = (-35/5) = -7
So, one point on the graph is (-10, -7).
Similarly, we can calculate the y-values for other x-values and plot the points. Once we have a few points, we can connect them to form the graph of the parabola.