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Which is the graph of the function f(x) = one-halfx2 + 2x – 6?

On a coordinate plane, a parabola opens up. It goes through (negative 6, 0), has a vertex at (negative 2, negative 8), and goes through (2, 0).
On a coordinate plane, a parabola opens up. It goes through (negative 2, 0), has a vertex at (2, negative 8), and goes through (6, 0).
On a coordinate plane, a parabola opens up. It goes through (negative 4, 6), has a vertex at (4, negative 10), and goes through (8, negative 6).
On a coordinate plane, a parabola opens up. It goes through (negative 8, negative 6), has a vertex at (negative 4, negative 10), and goes through (4, 6).

Respuesta :

calculista

Answer:

On a coordinate plane, a parabola opens up. It goes through (negative 6, 0), has a vertex at (negative 2, negative 8), and goes through (2, 0)

Step-by-step explanation:

step 1

Find the vertex

we have

[tex]f(x)=\frac{1}{2}x^{2}+2x-6[/tex]

This is a vertical parabola open upward (because the leading coefficient is positive)

The vertex represent a minimum

Convert the equation in vertex form

Factor 1/2

[tex]f(x)=\frac{1}{2}(x^{2}+4x)-6[/tex]

Complete the square

[tex]f(x)=\frac{1}{2}(x^{2}+4x+4)-6-2[/tex]

[tex]f(x)=\frac{1}{2}(x^{2}+4x+4)-8[/tex]

Rewrite as perfect squares

[tex]f(x)=\frac{1}{2}(x+2)^{2}-8[/tex]

The vertex is the point (-2,-8)

step 2

Find the x-intercepts

we know that

The x-intercepts are the values of x when the value of y is equal to zero

so

For f(x)=0

[tex]\frac{1}{2}(x+2)^{2}-8=0[/tex]

[tex]\frac{1}{2}(x+2)^{2}=8[/tex]

[tex](x+2)^{2}=16[/tex]

square root both sides

[tex]x+2=(+/-)4[/tex]

[tex]x=-2(+/-)4[/tex]

[tex]x=-2(+)4=2[/tex]

[tex]x=-2(-)4=-6[/tex]

so

The x-intercepts are the points (-6,0) and (2,0)

therefore

On a coordinate plane, a parabola opens up. It goes through (negative 6, 0), has a vertex at (negative 2, negative 8), and goes through (2, 0)

To get the graph of the given parabola, we will use the equation of the parabola to check if given points in options are lying on that parabola or not. If they lie on the graph of parabola,  then they would satisfy the equation, else not. We will use vertex form of parabola to get its vertex.

Thus, by below shown calculations, we have

Option A: On a coordinate plane, a parabola opens up. It goes through (-6, 0), has a vertex at (-2, -8), and goes through (2,0)

Given that:

The equation of the parabola is :

[tex]f(x) = \dfrac{1}{2}x^2 + 2x -6\\[/tex]

Calculations:

The equation can be rewritten in vertex form as:

[tex]f(x) = 0.5x^2 + 2x - 6\\f(x) = 0.5(x^2 + 4x) - 6 = 0.5(x^2 + 4x + 4 - 4) - 6\\f(x) = 0.5(x+2)^2 -8[/tex]

The vertex form [tex]f(x) = a(x-h)^2 + k[/tex] has vertex at (h,k)

Thus for the given parabola, the vertex is at (-2, -8)

If we put x = -6 and f(x) = 0 on LHS and RHS respectively, we get:

[tex]0 = 0.5\times (-6)^2 + 2 \times( -6) - 6 \\0 = 18 - 18\\0 = 0\\\rm LHS = \rm RHS[/tex]

Thus, point (-6,0) lies on the given parabola.

Similarly, if we put x = 2, f(x) = 0 in LHS and RHS respectively of the equation of parabola, we get:

[tex]0 = 0.5(2)^2 + 2 \times 2 - 6\\0 = 2 + 4 - 6\\0 = 0\\\rm LHS = \rm RHS[/tex]

Thus, point (2,0) too lies on the given parabola.

Thus, we have Option A: On a coordinate plane, a parabola opens up. It goes through (-6, 0), has a vertex at (-2, -8), and goes through (2,0)

as  the correct option.

The graph of the given parabola is attached below.

Learn more about parabola here:

https://brainly.com/question/8495268

Ver imagen astha8579

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