Sketching Parabolas help?

No, vertex is at ( 1/2 , -25/4), now a = 1 > 0 , which tells you the parabola opens up(U shaped)

to find the x-intercepts set y = 0:

(x - 1/2)^2 = 25/4

x - 1/2 = ± 5/2

x = 1/2 ± 5/2

x = -2 , 3

x-intercepts are at(-3 , 0) & (2 , 0)

to find the y-intercept set x = 0:

y = (0 -1/2)^2 - 25/4

= 1/4 - 25/4 = -24/4 = -6

y-intercept is at(0 , -6)

Now it should be an easy sketch.

Solve x^2 - x - 6 = 0 to get the points where the curve crosses the x-axis. You will get:

x = -2 and x = 3

Now we find the point at which the curve crosses the y axis, that is the value of y when x = 0. We get y = -6

Now you have four points which is sufficient to sketch the curve. Make sure that the curve is symmetrical about the vertex.

Also, label the points you have use with their co-ordinates. mark e the a xes and give the graph a title.

Sketching g is different from drawing. In the former we show the main features of the graph. Drawing means the curve has to be more exact. usually we would choose some values of x and calculate y. Then we have more points to draw the curve. You could do this for sketching, of course, but simply do not show these points

There are 3 thoughts to graph a parabola: a million. Plot the coordinates. it truly is the longest and toughest way, as there is too a lot room for blunders and some solutions may be fractional. 2. Vertex mode. positioned the quadratic equation into vertex style, plot the vertex, and use the -2,-a million,0,a million,2 technique, in case you comprehend a thanks to apply transformation of applications. 3. Conic section mode. this may require you in searching the vertex, directrix, and axis of symmetry to graph. This, alongside with vertex mode, calls for you to finish the sq.. without a issue to sparkling up, i will not help too a lot. i will in simple terms inform you thoughts to graph the parabola and so that you could use even with technique is ideal. omit Kristin