How do you graph the parabola #y=x^2-3x+8# using vertex, intercepts and additional points?

2 Answers

For #y=ax^2+bx+c# the #x# for vertex is

#x_v=-b/(2a)=3/2#

so

#y_v=x^2−3x+8 = (3/2)^2-3*3/2+8= 23/4 #

So the vertex is #( 3/2,23/4)#.

Since #a>0#, so it is looking up.
The #y#-intercept is #Y=0^2-3*0+8=8# so #(0,8)#

Explanation:

After finding the #y#-intercept, we should practically find the #x#-intercept, but as you see, the graph is looking upward since #a# is bigger than #0# and the vertex is higher than #0#.

So there will be no #x#-intercept, meaning that the graph will not intersect the #x#-axis. You can just plug in any number and then use the symmetrical characteristic of the hyperbola and draw the whole thing.

enter image source here

Dec 3, 2017

Vertex: #(\frac{3}{2},\frac{23}[4})#

#y#-intercept: #(0,8)#

Other point: #(3,8)#

Explanation:

We can find the #x#-coordinate of the vertex by using the formula #-\frac{b}{2a}#, where #a# and #b# come from the standard form of #ax^2+bx+c#.

In this equation,

  • #a=1#
  • #b=-3#
  • #c=8#

Plugging in yield:

#-\frac{-3}{2(1)}\quad\implies\quad \frac{3}{2}#

To find the #y#-coordinate of the vertex, plug the #x#-coordinate into the equation in standard form:

#x^2-3x+8#

#\implies (\frac{3}{2})^2-3(\frac{3}{2})+8#

#\implies \frac{9}{4}-\frac{9}{2}+8#

#\implies -\frac{9}{4}+8#

#\implies \frac{23}{4}#

#\therefore# the vertex is #(\frac{3}{2},\frac{23}{4})#.


The #y#-intercept is simply #(0,c)#, which is #(0,8)# in this case.


For a third point, we can find the axis of symmetry of the parabola, and reflect the #y#-intercept across that line.

The axis of symmetry is the vertex’s #x#-coordinate, which is #1.5#.

#\therefore# the third point will have an #x#-coordinate of #1.5\cdot 2#, which is #3#. The #y#-coordinate is the same as the #y-#-intercept, which is #8#.

#\therefore# a third point is #(3,8)#.

That’s all we need to graph a parabola.

Desmos