What is [tex]$f(x)=2 x^2+28 x-5$[/tex] written in vertex form?

A. [tex]$f(x)=2(x+7)^2-19$[/tex]
B. [tex][tex]$f(x)=2(x+7)^2-103$[/tex][/tex]
C. [tex]$f(x)=2(x+14)^2-14$[/tex]
D. [tex]$f(x)=2(x+14)^2-98$[/tex]



Answer :

To rewrite the quadratic function [tex]\( f(x) = 2x^2 + 28x - 5 \)[/tex] in vertex form, we need to complete the square. The standard form of a quadratic function is [tex]\( f(x) = ax^2 + bx + c \)[/tex], and the vertex form is [tex]\( f(x) = a(x - h)^2 + k \)[/tex], where [tex]\((h, k)\)[/tex] is the vertex of the parabola.

Here are the steps to transform the given quadratic function into vertex form:

1. Identify the coefficients: For the given function [tex]\( f(x) = 2x^2 + 28x - 5 \)[/tex],
[tex]\[ a = 2, \quad b = 28, \quad c = -5. \][/tex]

2. Find the vertex [tex]\(h\)[/tex]: The x-coordinate of the vertex [tex]\(h\)[/tex] is given by the formula
[tex]\[ h = -\frac{b}{2a}. \][/tex]
Plugging in the values of [tex]\(a\)[/tex] and [tex]\(b\)[/tex],
[tex]\[ h = -\frac{28}{2 \cdot 2} = -\frac{28}{4} = -7. \][/tex]

3. Find the vertex [tex]\(k\)[/tex]: The y-coordinate of the vertex [tex]\(k\)[/tex] can be found by substituting [tex]\(h\)[/tex] back into the function [tex]\(f(x)\)[/tex]:
[tex]\[ k = f(h) = f(-7). \][/tex]
Calculate [tex]\( f(-7) \)[/tex]:
[tex]\[ f(-7) = 2(-7)^2 + 28(-7) - 5. \][/tex]
Simplifying,
[tex]\[ f(-7) = 2(49) + (-196) - 5 = 98 - 196 - 5 = -98 - 5 = -103. \][/tex]

4. Rewrite the equation in vertex form: Now that we have [tex]\(h = -7\)[/tex] and [tex]\(k = -103\)[/tex], we can write the vertex form of the quadratic function:
[tex]\[ f(x) = 2(x + 7)^2 - 103. \][/tex]

Finally, comparing this result with the given options, we find that the correct vertex form of the function is:

[tex]\[ f(x) = 2(x + 7)^2 - 103. \][/tex]

Thus, the correct option is:

[tex]\[ f(x) = 2(x + 7)^2 - 103. \][/tex]