How would you solve the following equation by graphing 2x^2-13x-7=0?

Question

can you please teach me how to do it? im having such a hard time with that problem!

Answer 1

To solve this BY GRAPHING (which is what you asked for; a fact that has seemed to escape some of the other answerers), you'd start by graphing the function y = 2x^2 - 13x - 7.  Essentially, you want to find the values x that make = 0.  So look for the points on your graph where the graph crosses the x-axis.

Answer 2

use whatever program you have to generate the graph then locate all the spots where the graph crosses the x-axis.    

Answer 3

To solve a quadratic by graphing you need enough points to

1) Clearly show what kind of graph it is,

2) Show where the intercepts are located.

So just make a table of x and y and plug in a bunch of numbers.

First rewrite the equation in this format:

y = 2x^2-13x-7

I graphed this and found the following to suffice:

(0,-7), (-1,8), (-2,27), (-3,50), (1,-18), (2,-25), (3,-28), (4,-27), (5,-22), (6,-13), (7,0) (8,17), (9,38), (10,63)

Then you will notice that the intercepts are: (0,-7) and (7,0).

And as you most likely know the vertex is plotted using -b/2a, which in this case is:

-(-13) / 2(2) = 13/4

Using that value for x we punch it back into the original equation to get y and we get

2(13/4)^2 - 13(13/4) - 7 = y

-28.125 = y

So the vertex is located at (13/4, -28.125)

Have a good day!

Answer 4

To solve the equation by graphing

let us write it in the function form

y =2x^2-13x-7

get the co-ordinates by substituting different value of x and y in the equation and plot the graph . The graph where it intersects the x-axis, will be the solution of the equation since there is zero in RHS.

OR otherwise

write the equation in vertex form [ y = a(x-h)^2 +k]

y = 2 [ x^2 -13x/2 -7/2]

add and subtract  (1/2 *coeff of x )^2= (1/2 * -13/2)^2= 169/16

inside the bracket.

y = 2[ x^2 -13/2 +169/16) + (-7/2 - 169/16)]

y = 2 [(x-13/4)^2) - 281/16]

y = 2(x-13/4)^2 - 281/4

so,a= 2, h = 13/4, k= -281/4

so the vertex of the graph is ( 13/4, -281/4)

now , plot the graph with this vertex .

see where the graph intersects the x-axis, that will be the solution.

Answer 5

this is a quadratic equation in standard form (quadratic because it is squared, standard because it equals zero) saying the equation is equal to zero is the same thing as asking you to find the 2, 1, or 0 points at which the parabola (shape of the graph) intersect the x (horizontal) axis. to do this, it is easy to make an input (x) and output (y) table like so:

x      y

-2   27

-1   8

0   -7

1    -18

2   -25

3   -28

4 -27

the x values i just made up, and the y values i obtained by substituting the x into the equation. you can do this on a graphing calculator by going to Y= and then hitting 2nd graph (table). as you can see the lowest y value that the parabola will ever hit is -28. this is the minimum.

ANYWAY. now plot these points. in doing so you will get a parabola in which when y = 0 x = 7. that is your solution

Answer 6

2x^2-13x-7=0?

(2x + 1)(x - 7) = 0

2x + 1 = 0 -------> change sign on 1, then move to the other side

2x = -1 ----------> divided both side by 2

x = -1/2

x - 7 = 0 -------> change sign on -7, then move to the other side

x = 7