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.
