Question

HELPPPP PLEASEEE 45 POINTS

y=-2x^2+6x-5

1. What is the vertex?

2. Does it open up or down?

3. What is the intercept?

Answer 1

Answer:

Step-by-step explanation:

The first thing you should do is get a graph of this quadratic. The graph will answer the location of the vertex and it will also tell you if it opens up or down. It (finally) shows the intercept although that is easily found.

Question 3

You find the intercept by making x = 0

When you do that y becomes

$y = - 2(0)^2 + 6(0) - 5$

$y = 0 - 5$

$y = - 5$

Question 2

This too, is just a visual answer. Just look at the number in front of $x^2$

$y = ax^2$

if a < 0 then the graph always opens down

if a > 0 then the graph always opens up

In this case, $y = - 2x^2$. The graph opens down. The 6x + 5 does not affect the answer at all.

Question 1

This part is the tricky part and you have to complete the square.

Put brackets around the first 2 terms.

$y = ( - 2 \times x^2 + 6x ) - 5$    Pull out the common factor of - 2

$y = -2 (x^2 - 3x) - 5$        divide - 3 by 2 and square. Add inside the brackets.

$y = -2(x^2 - 3x + (\frac{3}{2})^2 ) - 5$   Add 2 times the squared amount outside the brackets.

$y = -2(x^2 - 3x + (\frac{3}{2})^2 ) - 5 + 2\times(\frac{3}{2} )^2$

$y = -2(x^2 - 3x+ \frac{9}{4})- 5 + \frac{9}{2}$     Show what is inside the brackets as a perfect square. Combine - 5 and 9/2

$y = -2(x - \frac{3}{2})^2 - 5 + 4.5\\y = -2(x - \frac{3}{2})^2 - 0.5$

The vertex should be at (3/2, - 0.5)

Answer: The graph confirms (3/2, - 0.5) as the vertex.