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Sagot :
Let's break down the solution to the problem step by step.
### Question 8
We are given the value of [tex]\( a \)[/tex] from question 7:
- [tex]\[ a = -0.1 \][/tex]
We need to write the quadratic equation in the form [tex]\( y = a(x - r_1)(x - r_2) \)[/tex]. For this problem, the roots (or [tex]\( x \)[/tex]-intercepts) of the quadratic equation are given as:
- [tex]\[ r_1 = 0 \][/tex]
- [tex]\[ r_2 = 50 \][/tex]
Substituting [tex]\( a \)[/tex], [tex]\( r_1 \)[/tex], and [tex]\( r_2 \)[/tex] into the quadratic form:
[tex]\[ y = -0.1(x - 0)(x - 50) \][/tex]
So, the equation in the form [tex]\( y = a(x - r_1)(x - r_2) \)[/tex] is:
[tex]\[ y = -0.1(x - 0)(x - 50) \][/tex]
### Question 9
Now, we need to use the distributive property to multiply the equation from question 8 to get the quadratic equation in the standard form [tex]\( y = ax^2 + bx \)[/tex].
Start with the equation we derived:
[tex]\[ y = -0.1(x)(x - 50) \][/tex]
- First, expand the product inside the parentheses:
[tex]\[ y = -0.1(x^2 - 50x) \][/tex]
- Next, distribute the [tex]\( -0.1 \)[/tex] across the terms inside the parentheses:
[tex]\[ y = -0.1x^2 + 5x \][/tex]
So, the quadratic equation in the standard form [tex]\( y = ax^2 + bx \)[/tex] is:
[tex]\[ y = -0.1x^2 + 5x \][/tex]
### Question 10
To create a graph of the parabola, you need the vertex and the [tex]\( x \)[/tex]-intercepts.
We've determined that the [tex]\( x \)[/tex]-intercepts are [tex]\( x = 0 \)[/tex] and [tex]\( x = 50 \)[/tex].
To find the vertex, we use the fact that for a parabola in the form [tex]\( y = ax^2 + bx \)[/tex]:
- The vertex [tex]\( x \)[/tex]-coordinate is given by [tex]\( x = -\frac{b}{2a} \)[/tex].
Here, [tex]\( a = -0.1 \)[/tex] and [tex]\( b = 5 \)[/tex].
[tex]\[ x = -\frac{5}{2(-0.1)} = -\frac{5}{-0.2} = 25 \][/tex]
Now, substitute [tex]\( x = 25 \)[/tex] back into the equation to find the [tex]\( y \)[/tex]-coordinate of the vertex:
[tex]\[ y = -0.1(25)^2 + 5(25) = -0.1(625) + 125 = -62.5 + 125 = 62.5 \][/tex]
So, the vertex is at [tex]\( (25, 62.5) \)[/tex].
With this information, you can create the graph of the parabola. The parabola will pass through the points [tex]\((0, 0)\)[/tex], [tex]\((50, 0)\)[/tex], and have the vertex at [tex]\( (25, 62.5) \)[/tex].
### Question 8
We are given the value of [tex]\( a \)[/tex] from question 7:
- [tex]\[ a = -0.1 \][/tex]
We need to write the quadratic equation in the form [tex]\( y = a(x - r_1)(x - r_2) \)[/tex]. For this problem, the roots (or [tex]\( x \)[/tex]-intercepts) of the quadratic equation are given as:
- [tex]\[ r_1 = 0 \][/tex]
- [tex]\[ r_2 = 50 \][/tex]
Substituting [tex]\( a \)[/tex], [tex]\( r_1 \)[/tex], and [tex]\( r_2 \)[/tex] into the quadratic form:
[tex]\[ y = -0.1(x - 0)(x - 50) \][/tex]
So, the equation in the form [tex]\( y = a(x - r_1)(x - r_2) \)[/tex] is:
[tex]\[ y = -0.1(x - 0)(x - 50) \][/tex]
### Question 9
Now, we need to use the distributive property to multiply the equation from question 8 to get the quadratic equation in the standard form [tex]\( y = ax^2 + bx \)[/tex].
Start with the equation we derived:
[tex]\[ y = -0.1(x)(x - 50) \][/tex]
- First, expand the product inside the parentheses:
[tex]\[ y = -0.1(x^2 - 50x) \][/tex]
- Next, distribute the [tex]\( -0.1 \)[/tex] across the terms inside the parentheses:
[tex]\[ y = -0.1x^2 + 5x \][/tex]
So, the quadratic equation in the standard form [tex]\( y = ax^2 + bx \)[/tex] is:
[tex]\[ y = -0.1x^2 + 5x \][/tex]
### Question 10
To create a graph of the parabola, you need the vertex and the [tex]\( x \)[/tex]-intercepts.
We've determined that the [tex]\( x \)[/tex]-intercepts are [tex]\( x = 0 \)[/tex] and [tex]\( x = 50 \)[/tex].
To find the vertex, we use the fact that for a parabola in the form [tex]\( y = ax^2 + bx \)[/tex]:
- The vertex [tex]\( x \)[/tex]-coordinate is given by [tex]\( x = -\frac{b}{2a} \)[/tex].
Here, [tex]\( a = -0.1 \)[/tex] and [tex]\( b = 5 \)[/tex].
[tex]\[ x = -\frac{5}{2(-0.1)} = -\frac{5}{-0.2} = 25 \][/tex]
Now, substitute [tex]\( x = 25 \)[/tex] back into the equation to find the [tex]\( y \)[/tex]-coordinate of the vertex:
[tex]\[ y = -0.1(25)^2 + 5(25) = -0.1(625) + 125 = -62.5 + 125 = 62.5 \][/tex]
So, the vertex is at [tex]\( (25, 62.5) \)[/tex].
With this information, you can create the graph of the parabola. The parabola will pass through the points [tex]\((0, 0)\)[/tex], [tex]\((50, 0)\)[/tex], and have the vertex at [tex]\( (25, 62.5) \)[/tex].
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