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A parabola graphed in the
xy-plane has equation
y=2x^(2)-5x-3. What is the
x-coordinate of the vertex of the parabola?

A parabola graphed in the $x y$ -plane has equation $y=2 x^{2}-5 x-3$ . What is the $x$ -coordinate of the vertex of the parabola?

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Write a quadratic function in vertex form

Full solution

Q. A parabola graphed in the

x y

-plane has equation

y=2 x^{2}-5 x-3

. What is the

x

-coordinate of the vertex of the parabola?

Identify quadratic equation: Identify the given quadratic equation. $\newline$ The given quadratic equation is $y = 2x^2 - 5x - 3$ .
Convert to vertex form: Convert the quadratic equation into vertex form. $\newline$ The vertex form of a quadratic equation is $y = a(x - h)^2 + k$ , where $(h, k)$ is the vertex of the parabola. $\newline$ To convert the given equation into vertex form, we need to complete the square.
Factor out $x^2$ coefficient: Factor out the coefficient of $x^2$ from the $x$ terms. $\newline$ The coefficient of $x^2$ is $2$ . We factor out $2$ from the $x$ terms to get: $\newline$ $y = 2(x^2 - (5/2)x) - 3$
Find $(b/2)^2$ : Find the value of $(b/2)^2$ to complete the square. $\newline$ The coefficient of $x$ is $-5/2$ . We need to find $(-5/4)^2$ to complete the square. $\newline$ $(-5/4)^2 = (25/16)$
Add and subtract $(b/2)^2$ : Add and subtract $(b/2)^2$ inside the parentheses and simplify. $\newline$ $y = 2(x^2 - \frac{5}{2}x + \frac{25}{16}) - 2*\frac{25}{16} - 3$ $\newline$ $y = 2(x - \frac{5}{4})^2 - \frac{50}{16} - 3$ $\newline$ $y = 2(x - \frac{5}{4})^2 - \frac{50}{16} - \frac{48}{16}$ $\newline$ $y = 2(x - \frac{5}{4})^2 - \frac{98}{16}$
Identify vertex x-coordinate: Identify the x-coordinate of the vertex. The x-coordinate of the vertex is the value of $h$ in the vertex form $y = a(x - h)^2 + k$ . From the equation $y = 2(x - \frac{5}{4})^2 - (\frac{98}{16})$ , we can see that $h = \frac{5}{4}$ .

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