bigaineQuAle6al bigaineQuAle6al
  • 08-11-2016
  • Mathematics
contestada

suppose a parabola has a vertex in quadrant IV and a<0 in the equation y=ax^2+bx+c. how many real solutions will the equation ax^2+bx+c=0 have?

Respuesta :

Edufirst
Edufirst Edufirst
  • 12-11-2016
Given the vertex is in quadrat IV, it is below the y-axis.

Given a < 0, the parabola open downdards.

Those two facts together implies that the parabola does not touch the y-axis, meaning y is never 0, and the equation ax^2 + bx + c =0 does not have any real solutions.

Answer; None

Answer Link

Otras preguntas

why did the spanish and portuguese launch voyages of exploration?
Humans get which of the following kinds of products from plants? a. medicines b. dyes c. rubber d. all of the above
how do you describe a republic government?
How many bits must be “flipped” (i.e., changed from 0 to 1 or from 1 to 0) in order to capitalize a lowercase ‘a’ that’s represented in ASCII?
factor the exspresion 8y^2-22y-21
How does meiosis affect genes
I need help on everything on 45 45 90 right triangles. I dont know how to solve two variables when one is shown.
What's the formula for a cylinder?
a football player gained 2 yards on one play. on the next play, he gained 5 feet. was his gain greater on the play or the second play? explain.
what is 1 6/7 -1 4/11