Use the factors of the function and the y-intercept to find the standard form of the equation representing Jared’s function. Type the correct answer in the box.
The Correct Answer and Explanation is:
To help you solve Part D of your problem, I need some clarification: You mention factors of the function and y-intercept, but you haven’t provided them. That information is crucial to generating the correct standard form of Jared’s function and writing the explanation.
However, I can give you a generic example showing how to write the standard form of a quadratic equation given its factors (x-intercepts) and y-intercept, which you can adapt to your specific problem.
Example Problem
Suppose Jared’s function has factors (x-intercepts) at x=2x = 2x=2 and x=5x = 5x=5, and the y-intercept is (0,10)(0, 10)(0,10).
Step 1: Write the factored form
The factors (x-intercepts) at x=2x = 2x=2 and x=5x = 5x=5 give the factored form: f(x)=a(x−2)(x−5)f(x) = a(x – 2)(x – 5)f(x)=a(x−2)(x−5)
Step 2: Use the y-intercept to find “a”
The y-intercept is the point where x=0x = 0x=0, and f(x)=10f(x) = 10f(x)=10. Plug into the equation: 10=a(0−2)(0−5)10=a(−2)(−5)10=a(10)a=110 = a(0 – 2)(0 – 5) \\ 10 = a(-2)(-5) \\ 10 = a(10) \\ a = 110=a(0−2)(0−5)10=a(−2)(−5)10=a(10)a=1
Step 3: Write the standard form
Now plug in a=1a = 1a=1 and expand the factored form: f(x)=(x−2)(x−5)=x2−5x−2x+10=x2−7x+10f(x) = (x – 2)(x – 5) \\ = x^2 – 5x – 2x + 10 \\ = x^2 – 7x + 10f(x)=(x−2)(x−5)=x2−5x−2x+10=x2−7x+10
So, the standard form is: f(x)=x2−7x+10f(x) = x^2 – 7x + 10f(x)=x2−7x+10
Explanation
To write a quadratic function in standard form, we start by understanding the information we’re given: the factors (also known as the x-intercepts or roots) and the y-intercept. The standard form of a quadratic function is: f(x)=ax2+bx+cf(x) = ax^2 + bx + cf(x)=ax2+bx+c
If you are given the roots (let’s say r1r_1r1 and r2r_2r2), you can first write the function in factored form: f(x)=a(x−r1)(x−r2)f(x) = a(x – r_1)(x – r_2)f(x)=a(x−r1)(x−r2)
This form is helpful because the roots directly give you the x-values where the graph crosses the x-axis.
However, to fully determine the equation, you need to know the value of “a” — the leading coefficient — which tells you how “wide” or “narrow” the parabola is, and whether it opens upward or downward. You can find “a” by plugging in a known point on the graph. The most common point used is the y-intercept, which is where x=0x = 0x=0.
By substituting the x and y values of the y-intercept into the factored form, you can solve for “a.” Once “a” is known, substitute it back into the equation. Then, expand the factored form using the distributive property to convert it into standard form. Simplify to get: f(x)=ax2+bx+cf(x) = ax^2 + bx + cf(x)=ax2+bx+c
This is the final answer — the function written in standard polynomial form.
