Question

question 1: consider the following function in vertex - form. convert it to standard form by expanding it and then identify the vertical intercept. ( g(x)=(x - 3)^{2}-4 ). standard form is ( g(x)=x^{2}-6x + 5 ) and the vertical intercept is 5. standard form is ( g(x)=x^{2}-6x + 1 ) and the vertical intercept is 1. none of the responses are correct. standard form is ( g(x)=x^{2}-4 ) and the vertical intercept is - 4. question 2: consider the following function in standard form. convert it to factored form by factoring, and identify the zeros. ( h(x)=x^{2}+3x - 10 ). factored form is ( h(x)=(x - 2)(x + 5) ) and the zeros are 2 and - 5. factored form is ( h(x)=(x - 2)(x + 5) ) and the zeros are - 2 and 5. none of the statements here is correct. factored form is ( h(x)=(x + 2)(x - 5) ) and the zeros are - 2 and 5. question 3: a parabola is symmetric, meaning that it is a mirror image of itself on either side of the vertex if you were to make a vertical fold going through the vertex. this means that the x - coordinate of the vertex has to lie exactly in the middle of the zeros of the parabola. for example, if the zeros were 1 and 3, then the x - coordinate of the vertex would be 2 since it is right in the middle of the zeros. note that taking the average of 1 and 3 is 2. now for this question suppose that the zeros of a function are - 4 and 8. which one of the following would be a possible vertex for the function? 0, 2 2, 0 0, 0

Explanation:

Question 1

Step1: Expand the vertex - form

Given $g(x)=(x - 8)^2-4$. Expand $(x - 8)^2$ using the formula $(a - b)^2=a^{2}-2ab + b^{2}$, where $a = x$ and $b = 8$. So $(x - 8)^2=x^{2}-16x + 64$. Then $g(x)=x^{2}-16x+64 - 4=x^{2}-16x + 60$. The vertex - form of a quadratic function is $y=a(x - h)^2+k$, and the vertex is $(h,k)$. Here $h = 8$ and $k=-4$, so the vertex is $(8,-4)$.
The standard form is $g(x)=x^{2}-16x + 60$ and the vertex is $(8,-4)$.

Step1: Factor the quadratic function

Given $h(x)=x^{2}+3x - 10$. We need to find two numbers that multiply to $-10$ and add up to $3$. The numbers are $5$ and $-2$ since $5\times(-2)=-10$ and $5+( - 2)=3$. So $h(x)=(x + 5)(x - 2)$.

Step2: Find the zeros

Set $h(x)=0$, then $(x + 5)(x - 2)=0$. By the zero - product property, $x+5 = 0$ or $x - 2=0$. Solving these equations gives $x=-5$ and $x = 2$.

Step1: Calculate the x - coordinate of the vertex

The zeros of the function are $-4$ and $6$. The x - coordinate of the vertex of a parabola is the average of the zeros. The formula for the average of two numbers $a$ and $b$ is $\frac{a + b}{2}$. Here $a=-4$ and $b = 6$, so $\frac{-4+6}{2}=\frac{2}{2}=1$.

Answer:

Standard form $g(x)=x^{2}-16x + 60$ and vertex is $(8,-4)$

Question 2