Solution
- The quadratic equation given is:
[tex]-x^2-4x+5=f(x)[/tex]Question 1:
- To know which direction a parabola opens, we follow the these rules:
[tex]\begin{gathered} \text{ Given,} \\ ax^2+bx+c \\ \\ \text{ if }a>0,\text{ the parabola opens Upwards} \\ \text{ If }a<0,\text{ the parabola opens downwards} \end{gathered}[/tex]- The value of a given is -1 < 0.
- Thus, the parabola opens downwards
Question 2:
- The y-intercept of the function is the coordinate of where the graph crosses the y-axis.
- In other words, this is also seen as the point where x = 0. Thus, substituting x = 0 into the equation given to us should readily reveal the y-value of the y-intercept.
- That is,
[tex]\begin{gathered} y=-x^2-4x+5 \\ put\text{ }x=0 \\ y=-0^2-4(0)+5 \\ y=5 \end{gathered}[/tex]- Thus, the y-intercept is (0, 5)
Question 3:
- The factorization of the function is given below:
[tex]\begin{gathered} f(x)=-x^2-4x+5 \\ \text{ We can rewrite the x-term as follows:} \\ -4x=-5x+x \\ \\ f(x)=-x^2-5x+x+5 \\ \text{ Thus, we can begin to factorize as follows:} \\ f(x)=-x(x+5)+1(x+5) \\ (x+5)\text{ is common so we can factorize again:} \\ \\ f(x)=(x+5)(1-x) \\ \\ f(x)=-1(x-1)(x+5) \end{gathered}[/tex]- Thus, the factorized form is:
[tex]f(x)=-1(x-1)(x+5)[/tex]Question 4:
[tex]\begin{gathered} \text{ The vertex of a parabola has its x-coordinate to be:} \\ x=-\frac{b}{2a}\text{ for the equation: }ax^2+bx+c \\ \\ \text{ Once we have the x-value, we can proceed to find the y-coordinate of the vertex by} \\ \text{ substituting this x-value into the equation.} \\ \text{ We have:} \\ \\ x=-\frac{-4}{2(-1)}=-2 \\ \\ \\ y=-x^2-4x+5 \\ put\text{ }x=-2 \\ y=-(-2)^2-4(-2)+5 \\ y=-4+8+5 \\ y=9 \end{gathered}[/tex]- Thus, the vertex of the parabola is (-2, 9)
Question 5:
- The vertex form of the parabola is given by:
[tex]\begin{gathered} \text{ The formula is:} \\ y=a(x-h)^2+k \\ where, \\ (h,k)\text{ is the coordinate of the vertex.} \\ \\ \text{ The question has already given us }a=-1,\text{ thus, we can find the equation of the vertex as follows:} \\ h=-2,k=9\text{ \lparen Gotten from question 4\rparen} \\ \\ \therefore y=-1(x-(-2))^2+9 \\ \\ y=-1(x+2)^2+9 \end{gathered}[/tex]- The equation of the vertex is
[tex]y=-1(x+2)^2+9[/tex]