Question

PLEASE HELP ME UNDERSTAND IDRK WHAT I DID WRONG I FOLLOWED MY TEACHERS NOTES

Answer 1

`\textsf{Vertex form:}\quad x=(1)/(8)(y-1)^2+4`

`\textsf{Standard form:}\quad x=(1)/(8)y^2-(1)/(4)y+(33)/(8)`

Explanation:

The graph shows a sideways parabola that opens to the right, with a vertex at (4, 1) and a focus at (6, 1).

The formula for a sideways parabola is:

`\large\boxed{(y-k)^2=4p(x-h)\quad \textsf{where $p\\eq 0$}}`

where:

• (h, k) is the vertex.
• (h+p, k) is the focus.

Since the vertex is (4, 1), then:

• h = 4
• k = 1

If the focus if (6, 1), then:

• h + p = 6
• k = 1

Find the value of p:

`\begin{aligned}h+p&=6\\4+p&=6\\p&=2\end{aligned}`

Substitute the values of h, k and p into the formula:

`\begin{aligned}(y-1)^2&=4(2)(x-4)\\(y-1)^2&=8(x-4)\end{aligned}`

Rearrange the equation to isolate x:

`\begin{aligned}(y-1)^2&=8(x-4)\\(1)/(8)(y-1)^2&=x-4\\x&=(1)/(8)(y-1)^2+4\end{aligned}`

Therefore, the equation of the parabola in vertex form is:

`\large\boxed{\boxed{x=(1)/(8)(y-1)^2+4}}`

The standard form of equation of a horizontal parabola is x = ay² + by + c. Therefore, to write the equation of the graphed parabola in standard for, expand and simplify the right side of the equation:

`\begin{aligned}x&=(1)/(8)(y-1)^2+4\\\\x&=(1)/(8)(y^2-2y+1)+4\\\\x&=(1)/(8)y^2-(1)/(4)y+(1)/(8)+4\\\\x&=(1)/(8)y^2-(1)/(4)y+(33)/(8)\end{aligned}`

Therefore, the equation of the parabola in standard form is:

`\large\boxed{\boxed{x=(1)/(8)y^2-(1)/(4)y+(33)/(8)}}`