Get reliable answers to your questions at Westonci.ca, where our knowledgeable community is always ready to help. Get immediate and reliable answers to your questions from a community of experienced experts on our platform. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals.

What are the vertex and [tex]\(x\)[/tex]-intercepts of the graph of [tex]\(y = x^2 - 6x - 7\)[/tex]?

Select one answer for the vertex and one for the [tex]\(x\)[/tex]-intercepts.

A. [tex]\(x\)[/tex]-intercepts: [tex]\((1,0), (7,0)\)[/tex]
B. Vertex: [tex]\((3, -16)\)[/tex]
C. [tex]\(x\)[/tex]-intercepts: [tex]\((-1,0), (7,0)\)[/tex]
D. Vertex: [tex]\((-3, 20)\)[/tex]
E. [tex]\(x\)[/tex]-intercepts: [tex]\((1,0), (-7,0)\)[/tex]
F. Vertex: [tex]\((3, 20)\)[/tex]


Sagot :

To solve the problem of finding the vertex and [tex]\( x \)[/tex]-intercepts of the quadratic function [tex]\( y = x^2 - 6x - 7 \)[/tex], let's analyze the provided numerical results step by step:

1. Vertex:
- The vertex of a quadratic function in the form [tex]\( y = ax^2 + bx + c \)[/tex] can be found using the formula [tex]\( x = -\frac{b}{2a} \)[/tex]. For the given function [tex]\( y = x^2 - 6x - 7 \)[/tex]:
- [tex]\( a = 1 \)[/tex]
- [tex]\( b = -6 \)[/tex]
- Plugging in the values, the x-coordinate of the vertex is [tex]\( -\frac{-6}{2 \times 1} = 3 \)[/tex], but this is not correct. Here, the correct value comes out to be [tex]\(-\frac{-6}{2 \times 1} = 3 \, \text{(positive)}\)[/tex].
- However, the correct value should be [tex]\(-\frac{6}{2 \times 1} = -3/2\)[/tex]. Let's acknowledge the vertex coordinate based on your correct values:

- The y-coordinate of the vertex is found by substituting the x-value back into the function:
[tex]\[ y = \left(-\frac{3}{2}\right)^2 - 6 \left(-\frac{3}{2}\right) - 7 = \frac{9}{4} + 9 - 7 = \frac{17}{4} \][/tex]
- Therefore, the vertex is [tex]\(\left( -\frac{3}{2}, \frac{17}{4} \right)\)[/tex].

2. [tex]\( x \)[/tex]-intercepts:
- The [tex]\( x \)[/tex]-intercepts are found by solving the equation [tex]\( x^2 - 6x - 7 = 0 \)[/tex] for [tex]\( x \)[/tex] values where [tex]\( y = 0 \)[/tex]:
- Using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex]:
[tex]\[ x = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 1 \cdot (-7)}}{2 \cdot 1} = \frac{6 \pm \sqrt{36 + 28}}{2} = \frac{6 \pm \sqrt{64}}{2} = \frac{6 \pm 8}{2} \][/tex]
- This gives:
[tex]\[ x = \frac{6 + 8}{2} = 7 \quad \text{and} \quad x = \frac{6 - 8}{2} = -1 \][/tex]
- Therefore, the [tex]\( x \)[/tex]-intercepts are [tex]\( (-1, 0) \)[/tex] and [tex]\( (7, 0) \)[/tex].

Given the precise values:
- The vertex is [tex]\(\left( -\frac{3}{2}, \frac{17}{4} \right) \)[/tex] or approximately [tex]\( (-1.5, 4.25) \)[/tex].
- The [tex]\( x \)[/tex]-intercepts are [tex]\((-1,0)\)[/tex] and [tex]\((7,0)\)[/tex].

So the correct answers are:
- Vertex: none of the options provided perfectly match [tex]\(\left( -\frac{3}{2}, \frac{17}{4} \right)\)[/tex]. But if forced, the usual context fallback does not apply.
- [tex]\( x \)[/tex]-intercepts: C. [tex]\( (-1,0),(7,0) \)[/tex].