Question

Standard form of the quadratic is y = __x^2 + __x +__.
On the graph it opens upwards, the vertex is (3,-1) and it crosses the y at 8

Answer 1

To find the standard form of the quadratic \(y = ax^2 + bx + c\) given the vertex and the y-intercept, we start with the vertex form of a quadratic equation, which is:

\[
y = a(x - h)^2 + k
\]

where \((h, k)\) is the vertex of the parabola. In this case, the vertex is \((3, -1)\), so:

\[
y = a(x - 3)^2 - 1
\]

Next, we know that the quadratic crosses the y-axis at \(y = 8\). The y-intercept occurs when \(x = 0\). We can substitute \(x = 0\) into the vertex form to find \(a\):

\[
8 = a(0 - 3)^2 - 1
\]

This simplifies to:

\[
8 = 9a - 1
\]

Solving for \(a\):

\[
9a = 8 + 1
\]
\[
9a = 9
\]
\[
a = 1
\]

Now we can substitute \(a\) back into the vertex form:

\[
y = 1(x - 3)^2 - 1
\]

Expanding this:

\[
y = (x - 3)^2 - 1
\]
\[
y = (x^2 - 6x + 9) - 1
\]
\[
y = x^2 - 6x + 8
\]

Thus, the standard form of the quadratic is:

\[
y = 1x^2 - 6x + 8
\]

So, filling in the blanks:

\[
y = 1x^2 + (-6)x + 8
\]