Question

a quadratic function $f(x)$ is hidden from view. you must find the $y$-intercept(s) of $f(x)$ and write the answer(s) in the form $(x, y)$. choose the form of the quadratic function $f(x)$ that you would like to see in order to answer the question most efficiently.

form: select a form

answer attempt 1 out of 2

there are no $y$-intercepts

submit answer

Explanation:

Step1: Recall y - intercept definition

The y - intercept of a function \(y = f(x)\) is the point where \(x = 0\), so we substitute \(x = 0\) into the function. For a quadratic function, the standard form is \(f(x)=ax^{2}+bx + c\) (where \(a
eq0\)). When \(x = 0\), \(f(0)=a(0)^{2}+b(0)+c=c\). So the y - intercept is \((0,c)\). The vertex form is \(f(x)=a(x - h)^{2}+k\), and when \(x = 0\), \(f(0)=a(0 - h)^{2}+k=ah^{2}+k\), which requires more calculation. The factored form is \(f(x)=a(x - r_1)(x - r_2)\), and when \(x = 0\), \(f(0)=a(-r_1)(-r_2)=ar_1r_2\), also more calculation. So the standard form \(f(x)=ax^{2}+bx + c\) is the most efficient to find the y - intercept because we can directly read \(c\) and the y - intercept is \((0,c)\).

Step2: Determine the y - intercept

Since every quadratic function (a polynomial of degree 2) will intersect the y - axis at \(x = 0\) (because when \(x = 0\), we can always compute \(f(0)\) as a real number for real - valued coefficients). The statement "There are no y - intercepts" is incorrect. For a quadratic function \(f(x)=ax^{2}+bx + c\), when \(x = 0\), \(y=c\), so the y - intercept is \((0,c)\). But since we don't have the specific function, if we assume the function is in standard form \(f(x)=ax^{2}+bx + c\), the y - intercept is found by setting \(x = 0\), so \(y = f(0)=c\), and the point is \((0,c)\).

Answer:

To find the y - intercept most efficiently, we choose the standard form \(f(x)=ax^{2}+bx + c\). The y - intercept of a quadratic function \(f(x)\) (which always exists) is \((0,c)\) where \(c\) is the constant term in the standard form. So the correct y - intercept exists and is of the form \((0,c)\) (for the standard form \(f(x)=ax^{2}+bx + c\)).