Answer :
To model the situation described by the quadratic equation [tex]\(y = a(x-h)^2 + k\)[/tex], we need to determine the values of [tex]\(a\)[/tex], [tex]\(h\)[/tex], and [tex]\(k\)[/tex]. We are provided with the following information:
- The lowest point (vertex) of the cable is [tex]\(6\)[/tex] feet above the roadway and is [tex]\(90\)[/tex] feet horizontally from the left bridge support. This gives us the vertex [tex]\((h, k)\)[/tex], where [tex]\(h = 90\)[/tex] and [tex]\(k = 6\)[/tex].
- At a horizontal distance of [tex]\(30\)[/tex] feet, the height of the cable is [tex]\(15\)[/tex] feet above the roadway.
Given these values, the vertex form of our equation is:
[tex]\[ y = a(x - 90)^2 + 6 \][/tex]
Next, we use the information that at [tex]\(x = 30\)[/tex], [tex]\(y = 15\)[/tex] to find the value of [tex]\(a\)[/tex]. Substitute [tex]\(x = 30\)[/tex] and [tex]\(y = 15\)[/tex] into the equation:
[tex]\[ 15 = a(30 - 90)^2 + 6 \][/tex]
This simplifies to:
[tex]\[ 15 = a(-60)^2 + 6 \][/tex]
[tex]\[ 15 = 3600a + 6 \][/tex]
To isolate [tex]\(a\)[/tex], subtract [tex]\(6\)[/tex] from both sides:
[tex]\[ 9 = 3600a \][/tex]
Finally, solve for [tex]\(a\)[/tex]:
[tex]\[ a = \frac{9}{3600} = \frac{1}{400} \][/tex]
So, the quadratic equation that models the situation is:
[tex]\[ y = \frac{1}{400}(x - 90)^2 + 6 \][/tex]
Therefore, the correctly modeled quadratic equation is:
[tex]\[ y = \frac{1}{400}(x - 90)^2 + 6 \][/tex]
- The lowest point (vertex) of the cable is [tex]\(6\)[/tex] feet above the roadway and is [tex]\(90\)[/tex] feet horizontally from the left bridge support. This gives us the vertex [tex]\((h, k)\)[/tex], where [tex]\(h = 90\)[/tex] and [tex]\(k = 6\)[/tex].
- At a horizontal distance of [tex]\(30\)[/tex] feet, the height of the cable is [tex]\(15\)[/tex] feet above the roadway.
Given these values, the vertex form of our equation is:
[tex]\[ y = a(x - 90)^2 + 6 \][/tex]
Next, we use the information that at [tex]\(x = 30\)[/tex], [tex]\(y = 15\)[/tex] to find the value of [tex]\(a\)[/tex]. Substitute [tex]\(x = 30\)[/tex] and [tex]\(y = 15\)[/tex] into the equation:
[tex]\[ 15 = a(30 - 90)^2 + 6 \][/tex]
This simplifies to:
[tex]\[ 15 = a(-60)^2 + 6 \][/tex]
[tex]\[ 15 = 3600a + 6 \][/tex]
To isolate [tex]\(a\)[/tex], subtract [tex]\(6\)[/tex] from both sides:
[tex]\[ 9 = 3600a \][/tex]
Finally, solve for [tex]\(a\)[/tex]:
[tex]\[ a = \frac{9}{3600} = \frac{1}{400} \][/tex]
So, the quadratic equation that models the situation is:
[tex]\[ y = \frac{1}{400}(x - 90)^2 + 6 \][/tex]
Therefore, the correctly modeled quadratic equation is:
[tex]\[ y = \frac{1}{400}(x - 90)^2 + 6 \][/tex]
