xuiene
Answered

I have to find two additional points, but can't seem to figure out how to find them even with the many videos ive watched. The equation is y = x2 + 6x – 7. I have the vertex, y-intercept & x-intercept, and axis, but i just can't seem to understand how to find those last two points i need pls help :(



Answer :

Step-by-step explanation:

To find 2 additional points to complete the graph, we can pick any [tex]x[/tex] value that is quite far from the known points.

The common points to draw a parabolic graph are:

  • x-intercept
  • y-intercept
  • vertex

To find the x-intercept, where [tex]y=0[/tex], we substitute the [tex]y[/tex] with 0:

[tex]y=x^2+6x-7[/tex]

[tex]0=x^2+6x-7[/tex]

[tex](x+7)(x-1)=0[/tex]

[tex]x=-7\ or\ 1[/tex]

Hence, x-intercept = [tex](-7,0)[/tex] and [tex](1,0)[/tex]

To find the y-intercept, where [tex]x=0[/tex], we substitute the [tex]x[/tex] with 0:

[tex]y=x^2+6x-7[/tex]

[tex]y=(0)^2+6(0)-7[/tex]

[tex]y=-7[/tex]

Hence, y-intercept = [tex](0,-7)[/tex]

To find the vertex, we can use this formula:

[tex]\displaystyle (x,y)=\left(-\frac{b}{2a} ,-\frac{b^2-4ac}{4a} \right)[/tex]

         [tex]\displaystyle =\left(-\frac{6}{2(1)} ,-\frac{6^2-4(1)(-7)}{4(1)} \right)[/tex]

         [tex]=(-3,-16)[/tex]

Since the above known points have [tex]x-value[/tex] of -7, -3 and 1, we pick:

  • 1 x-value that is reasonably less than -7
  • 1 x-value that is reasonably more than 1

to complete the graph.

Let's say we pick [tex]x=-10[/tex] and [tex]x=5[/tex], then we substitute both x-value into the equation to find the points.

x₁ = -10

[tex]y_1=x_1^2+6x_1-7[/tex]

    [tex]=(-10)^2+6(-10)-7[/tex]

    [tex]=33[/tex]

Hence, point 1 = (-10, 33)

x₂ = 5

[tex]y_2=x_2^2+6x_2-7[/tex]

    [tex]=(5)^2+6(5)-7[/tex]

    [tex]=48[/tex]

Hence, point 2 = (5, 48)

Notes:

If the [tex]y-values[/tex] of the points are beyond the provided space, you can pick other [tex]x-values[/tex] that are closer to the known points.