Question

PLEASE HURRY Which equation could generate the curve in the graph below? On a coordinate plane, a parabola is in quadrant 2 and opens up. The vertex is on the x-axis. y = 9x2 + 6x + 4 y = 6x2 – 12x – 6 y = 3x2 + 7x + 5 y = 2x2 + 8x + 8

Answer 1

Answer:

is D

Step-by-step explanation:

Answer 2

The equation that could represent the graph of the parabola is (d) $y = 2x^2 + 8x + 8$

From the question, we have the following highlights

• The parabola opens up
• The vertex is on the x-axis

The highlights above mean that: the y-coordinate (k) of the vertex is positive.

For a parabola: $y = ax^2 + bx + c$, the vertex is:

$(h,k) = (-\frac{b}{2a},f(h))$

Next, we test the options

(a) y = 9x2 + 6x + 4

We have:

$h = -\frac{6}{2 \times 9}$

$h = -\frac{1}{3}$

Substitute -1/3 for x in the function

$y = 9x^2 + 6x + 4$

$k = 9(1/3)^2 + 6(1/3) + 4$

$k = 7$

The value of k is not zero.

So, this cannot represent the parabola

(b) y = 6x2 – 12x – 6

We have:

$h = -\frac{-12}{2\times 6}$

$h = 1$

Substitute 1 for x in the function

$y = 6x^2 - 12x - 6$

$k = 6(1)^2 - 12(1) - 6$

$k = -12$

The value of k is not zero.

So, this cannot represent the parabola

(c) y = 3x2 + 7x + 5

We have:

$h = -\frac{7}{2\times 3}$

$h = -\frac{7}{6}$

Substitute -7/6 for x in the function

$y = 3x^2 + 7x + 5$

$k = 3(-7/6)^2 + 7(-7/6) + 5$

$k = 0.92$

The value of k is not zero.

So, this cannot represent the parabola

(b) y = 2x2 + 8x + 8

We have:

$h = -\frac{8}{2\times 2}$

$h = -2$

Substitute -2 for x in the function

$y = 2x^2 + 8x + 8$

$k = 2(-2)^2 + 8(-2) + 8$

$k = 0$

The y-coordinate (k) is zero.

Hence, $y = 2x^2 + 8x + 8$ could represent the parabola

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