Write x2 - 2x + 5 = 0 in vertex form. (Be sure to write your response in the form a(x-h)^2+k=0 with no spaces!)

1 answer:

4 0

To write the given quadratic equation to its vertex form, we first form a perfect square.

x² - 2x + 5 = 0
Transpose the constant to other side of the equation,
x² - 2x = -5
Complete the square in the left side of the equation,
x² - 2x + (-2/1(2))² = -5 + (-2/1(2))²
Performed the operation,
x² - 2x + 1 = -5 + 1
Factor the left side of the equation,
(x - 1)² = -4

Thus, the vertex form of the equation is,
<em> (x-1)² + 4 = 0</em>

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Find the standard form of the equation of the parabola with a vertex at the origin and a focus at (0, -7).

creativ13 [48]

So hmm check the picture below

we know the vertex is at the origin, and the focus point is below it, that means two things, the parabola is vertical and it's opening downwards

notice the distance "p", from the vertex to the focus point, is just 7 units, however, since the parabola is opening downwards, the "p" value will be negative, so p = -7

$\bf \textit{parabola vertex form with focus point distance}\\\\
\begin{array}{llll}
(y-{{ k}})^2=4{{ p}}(x-{{ h}}) \\\\
\boxed{4{{ p}}(y-{{ k}})=(x-{{ h}})^2 }\\
\end{array}
\qquad 
\begin{array}{llll}
vertex\ ({{ h}},{{ k}})\\\\
{{ p}}=\textit{distance from vertex to }\\
\qquad \textit{ focus or directrix}
\end{array}\\\\
-------------------------------\\\\
4{{ p}}(y-{{ k}})=(x-{{ h}})^2\quad 
\begin{cases}
h=0\\
k=0\\
p=-7
\end{cases}\implies 4(-7)(y-0)=(x-0)^2
\\\\\\
-28y=x^2\implies y=-\cfrac{1}{28}x^2$