Respuesta :
Remember that a quadratic with two real zeroes can be written as [tex]a(x - r_1)(x - r_2)[/tex], where [tex]a[/tex] is a constant and [tex]r_1[/tex] and [tex]r_2[/tex] are the zeroes (or roots) of the function. Since the graph shows that the two zeroes are at -6 and 6, the equation has to be of the form
[tex]y = a(x - ({-6}))(x - 6)[/tex], or
[tex]y = a(x + 6)(x - 6)[/tex]
To solve for a, let's use the point at the vertex (0, 36) and plug that in:
[tex]36 = a(0 + 6)(0 - 6)[/tex]
[tex]36 = {-36}a[/tex]
[tex]a = {-1}[/tex]
(It makes sense that [tex]a[/tex] is negative since the parabola opens down.)
So, the equation of the parabola is
[tex]y = -(x + 6)(x - 6)[/tex], or
[tex]\bf y = -x^2 + 36[/tex]
Now for the second part, just pick any two points with which we can draw a line with a positive slope. I'll use x = -2 and 1:
[tex]y = -({-2})^2 + 36 = {-4} + 36 = 32[/tex]
[tex]y = -(1)^2 + 36 = {-1} + 36 = 35[/tex]
So, our two points are (-2, 32) and (1, 35). To find the equation of the linear function that goes through these two points, let's use slope-intercept form, which is [tex]f(x) = mx + b[/tex]. The slope [tex]m[/tex] is given by [tex]\frac{y_2 - y_1}{x_2 - x_1}[/tex], so
[tex]m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{35 - 32}{1 - ({-2})} = 1[/tex]
So, the equation of the linear function so far is just [tex]f(x) = x + b[/tex], and we can find [tex]b[/tex] by plugging in one of the points on the line:
[tex]35 = 1 + b[/tex]
[tex]b = 34[/tex]
Thus, the equation of the linear function is
[tex]\bf f(x) = x + 34[/tex]
And you can find more points on the line simply by plugging other values of x, such as (0, 34) and (5, 39).
[tex]y = a(x - ({-6}))(x - 6)[/tex], or
[tex]y = a(x + 6)(x - 6)[/tex]
To solve for a, let's use the point at the vertex (0, 36) and plug that in:
[tex]36 = a(0 + 6)(0 - 6)[/tex]
[tex]36 = {-36}a[/tex]
[tex]a = {-1}[/tex]
(It makes sense that [tex]a[/tex] is negative since the parabola opens down.)
So, the equation of the parabola is
[tex]y = -(x + 6)(x - 6)[/tex], or
[tex]\bf y = -x^2 + 36[/tex]
Now for the second part, just pick any two points with which we can draw a line with a positive slope. I'll use x = -2 and 1:
[tex]y = -({-2})^2 + 36 = {-4} + 36 = 32[/tex]
[tex]y = -(1)^2 + 36 = {-1} + 36 = 35[/tex]
So, our two points are (-2, 32) and (1, 35). To find the equation of the linear function that goes through these two points, let's use slope-intercept form, which is [tex]f(x) = mx + b[/tex]. The slope [tex]m[/tex] is given by [tex]\frac{y_2 - y_1}{x_2 - x_1}[/tex], so
[tex]m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{35 - 32}{1 - ({-2})} = 1[/tex]
So, the equation of the linear function so far is just [tex]f(x) = x + b[/tex], and we can find [tex]b[/tex] by plugging in one of the points on the line:
[tex]35 = 1 + b[/tex]
[tex]b = 34[/tex]
Thus, the equation of the linear function is
[tex]\bf f(x) = x + 34[/tex]
And you can find more points on the line simply by plugging other values of x, such as (0, 34) and (5, 39).