Question

1. from chapter 3.2: what is the vertex of this quadratic function? type your answer as an ordered pair.

write all work on the accompanying worksheet.
$y = -3(x + 2)^2 + 4$
your answer

1. from chapter 3.2: what is the y-intercept of this quadratic function?

write all work on the accompanying worksheet.
$f(x) = x^2 + 3x - 5$
\\(\circ\\) (3, 5)
\\(\circ\\) (5, 0)

Explanation:

Question 6

Step1: Recall vertex form of quadratic

The vertex form of a quadratic function is \( y = a(x - h)^2 + k \), where \((h, k)\) is the vertex.

Step2: Identify h and k from given equation

Given \( y = -3(x + 2)^2 + 4 \), we can rewrite \( (x + 2) \) as \( (x - (-2)) \). So comparing with \( y = a(x - h)^2 + k \), we have \( h = -2 \) and \( k = 4 \).

Step1: Recall y - intercept definition

The y - intercept of a function \( y = f(x) \) is the value of \( y \) when \( x = 0 \).

Step2: Substitute x = 0 into the function

For \( f(x)=x^{2}+3x - 5 \), when \( x = 0 \), we have \( f(0)=0^{2}+3(0)-5=- 5 \). So the y - intercept is the point \((0,-5)\). But looking at the given options, there seems to be a mistake in the options provided. However, if we follow the calculation:
\( f(0)=0 + 0-5=-5 \), so the y - intercept is \((0, - 5)\). But since the options given are \((3,5)\) and \((5,0)\) which are incorrect, but if we assume there is a typo and maybe the function was different, but based on the given function \( f(x)=x^{2}+3x - 5 \), the y - intercept is \((0,-5)\). But since the options are not correct, but if we have to choose from the given options, none of them is correct. But if we made a mistake in the problem understanding, let's re - check. The y - intercept is found by \( x = 0 \), so \( f(0)=0 + 0-5=-5\), so the point is \((0,-5)\).

Answer:

\((-2, 4)\)

Question 7