Question

2.3 writing quadratic functions (pp. 141 - 150) learning target: write equations of quadratic functions using given characteristics. the table or graph represents a quadratic function. write an equation of the function in standard form. 17. x -5 -4 -3 -2 g(x) 5 2 5 14 18.

Explanation:

Step1: Recall quadratic function standard form

The standard - form of a quadratic function is \(y = ax^{2}+bx + c\).

Step2: For problem 17

Substitute the points \((-5,5)\), \((-4,2)\), \((-3,5)\) into \(y = ax^{2}+bx + c\):
When \(x=-5\) and \(y = 5\), we have \(5=a(-5)^{2}+b(-5)+c=25a-5b + c\).
When \(x=-4\) and \(y = 2\), we have \(2=a(-4)^{2}+b(-4)+c=16a-4b + c\).
When \(x=-3\) and \(y = 5\), we have \(5=a(-3)^{2}+b(-3)+c=9a-3b + c\).
Subtract the second equation from the first equation: \((25a-5b + c)-(16a-4b + c)=5 - 2\), which simplifies to \(9a - b=3\).
Subtract the third equation from the second equation: \((16a-4b + c)-(9a-3b + c)=2 - 5\), which simplifies to \(7a - b=-3\).
Subtract the new - obtained second equation from the new - obtained first equation: \((9a - b)-(7a - b)=3-(-3)\), \(2a = 6\), so \(a = 3\).
Substitute \(a = 3\) into \(9a - b=3\), we get \(9\times3 - b=3\), \(27 - b=3\), \(b = 24\).
Substitute \(a = 3\) and \(b = 24\) into \(25a-5b + c=5\), \(25\times3-5\times24 + c=5\), \(75-120 + c=5\), \(c = 50\). So \(g(x)=3x^{2}+24x + 50\).

Step3: For problem 18

Since the quadratic function has \(x\) - intercepts at \(x = 2\) and \(x = 10\), the factored form is \(y=a(x - 2)(x - 10)\).
Substitute the point \((6,-4)\) into \(y=a(x - 2)(x - 10)\): \(-4=a(6 - 2)(6 - 10)\), \(-4=a\times4\times(-4)\), \(-4=-16a\), \(a=\frac{1}{4}\).
Expand \(y=\frac{1}{4}(x - 2)(x - 10)=\frac{1}{4}(x^{2}-10x-2x + 20)=\frac{1}{4}x^{2}-3x + 5\).

Answer:

1. \(g(x)=3x^{2}+24x + 50\)
2. \(y=\frac{1}{4}x^{2}-3x + 5\)