How do you graph the parabola f(x)=x ^2 + 3x-10 using vertex, intercepts and additional points?

2 Answers
Jan 11, 2018

After calculating the vertex and intercepts and plotting them, calculate additional points, (x,f(x)), near these crucial spots of our parabola.

Explanation:

To find the vertex, we use the formula x = -b/(2a), but as you may have guessed, this will only give us the x coordinate of the vertex. In order to get the y coordinate, we would need to put our x value back into f(x).

Standard form is
ax^2 + bx - 10

In our case:
a = 1
b = 3

This means the x coordinate of our vertex will be

-(3)/(2(1)) or simply -3/2

Plugging the x coordinate back into f(x), we get the y coordinate.

(-3/2)^2+3(-3/2)-10

= 9/4-9/2-10

= -49/4

Our vertex is therefore (-3/2,-49/4).

Now, we can calculate the x-intercept and the y-intercept, which, alongside our vertex, will both help us determine what other nearby points we may wish to calculate.

The x-intercept is where y=0. This case may be solved by factoring.

x^2 + 3x -10 = 0

(x + 5)(x - 2) = 0

x + 5 = 0 or x - 2 = 0

x = - 5 or x = 2

So, our x-intercepts are (-5,0) and (2,0).

The y-intercept is where x=0. Simply plug this value into f(x).

(0)^2+3(0)-10

= -10

So, our y-intercept is (0,-10).

Now, we have four relevant points that we can plot. These are:

  • The vertex, (-3/2,-49/4)
  • The x-intercepts, (-5,0) and (2,0)
  • And the y-intercept, (0,-10)

After plotting these, we likely wish to plot some additional points in between to more accurately graph it. We would choose x values and plug them into f(x).

These would be points within our range of crucial coordinates, so anywhere between our lowest x coordinate, (-5,0), and our greatest x coordinate, (2,0).

You may also plot more points outside of this range to give a more complete picture.

Jan 11, 2018

ME (Harold Walden)ME (Harold Walden)

Explanation:

I have handwritten my explanation, I hope it is legible :)Again, the source is me :)Again, the source is me :)

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