Question

Please help me with 6-7 / 10 points

Answer 1

Answer:

6) Axis of symmetry x=1. Vertex at (5,1)

7) Axis of symmetry x=1. Vertex at (1,2). y-intercept at y = -1

Step-by-step explanation:

There is an equation to find the axis of symmetry of a quadratic function and is given by:

Knowing that the standard form of a quadratic equation is:

$y=ax^{2}+bx+c$

The axis of symmetry is:

$x=\frac{-b}{2a}$ (1)

(6)

The quadratic equation of the first problem is:

$y=2x^{2}-4x+7$

Then, using equation (1) the axis of symmetry will be:

$x=\frac{-(-4)}{2(2)}$

$x=\frac{4)}{4}$

$x=1$      

We need to use this value of x to find the vertex;

$y(1)=2(1)^{2}-4(1)+7$

$y(1)=5$

The vertex is (5,1)

Therefore, we disagree with Ahmed. The axis of symmetry is x = 1 and the vertex is in the point (5,1).

(7)

We can use the same method here. The function is:

$f(x)=-3x^{2}+6x-1$

The axis of symmetry is:

$x=\frac{-6}{2(-3)}$

$x=1$

The vertex is:

$f(1)=-3(1)^{2}+6(1)-1$

$f(1)=2$

So the vertex is (1,2)

We need to evaluate the function at x = 0 to get the y-intercept value.

$f(0)=-3(0)^{2}+6(0)-1$

$f(0)=-1$

Therefore, the y-intercept is at y = -1.

I hope it helps you!