Question

4.5 introduction to modeling
question 2 of 3 (3 points) | question attempt: 1 of unlimited
vertex: (3,4)
(c) find the equation of the axis of symmetry.
equation of axis of symmetry: x = 3
(d) find the intercept(s).
for both the x - and y - intercept(s), make sure to do the following.

• if there is more than one, separate them with commas.
• if there are none, select

one\.
x - intercept(s):
y - intercept(s):

Explanation:

Step1: Find x - intercept

Set \(y = 0\). For a quadratic function in vertex - form \(y=a(x - h)^2+k\) (where \((h,k)=(3,4)\)), we have \(0=a(x - 3)^2+4\). Then \(a(x - 3)^2=-4\). If \(a
eq0\), \((x - 3)^2=-\frac{4}{a}\). Since the square of a real number \(x-3\) is non - negative (\((x - 3)^2\geq0\)) and \(-\frac{4}{a}<0\) for \(a
eq0\), there are no real solutions for \(x\). So the \(x\) - intercepts are None.

Step2: Find y - intercept

Set \(x = 0\). For a quadratic function \(y=a(x - 3)^2+4\), when \(x = 0\), \(y=a(0 - 3)^2+4=9a + 4\). Without knowing the value of \(a\), we assume the general form. If the function is a quadratic with vertex \((3,4)\), when \(x = 0\), \(y\) is a single value. Substituting \(x = 0\) into the general form of the quadratic function related to the vertex \((h,k)\): \(y=a(x - h)^2+k\) gives \(y=a(0 - 3)^2+4=9a+4\). When \(a = 1\) (assuming a standard form for simplicity, since the shape of finding the \(y\) - intercept is the same for non - zero \(a\)), \(y=9\times1 + 4=13\). The \(y\) - intercept is the point \((0,9a + 4)\) or when considering the general case of a quadratic with the given vertex, the \(y\) - intercept is \(13\) (assuming \(a = 1\)). In general, for a quadratic function with vertex \((3,4)\) in the form \(y=a(x - 3)^2+4\), the \(y\) - intercept is the value of \(y\) when \(x = 0\), which is \(9a+4\).

Answer:

x - intercept(s): None
y - intercept(s): \(13\) (assuming \(a = 1\) in \(y=a(x - 3)^2+4\))