Question

consider the function $f(x) = -2x^2 + 12x - 10$. the y-intercept of $f(x)$ is $(0, a)$, where $a = \square$. the x-intercepts of $f(x)$ are $(b, 0)$ and $(c, 0)$, where $b = \square$ and $c = \square$. lastly, the axis of symmetry of $f(x)$ is $x = d$, where $d = \square$. note: your answers should be integers.

Explanation:

Step1: Find y-intercept (A)

To find the y-intercept, we set \( x = 0 \) in the function \( f(x)=-2x^{2}+12x - 10 \).
Substitute \( x = 0 \): \( f(0)=-2(0)^{2}+12(0)-10=- 10 \). So \( A=-10 \).

Step2: Find x-intercepts (B and C)

To find the x-intercepts, we set \( f(x) = 0 \), so we solve the quadratic equation \( -2x^{2}+12x - 10=0 \).
First, divide the entire equation by - 2 to simplify: \( x^{2}-6x + 5=0 \).
Factor the quadratic: \( x^{2}-6x + 5=(x - 1)(x - 5)=0 \).
Set each factor equal to zero: \( x-1 = 0\) gives \( x = 1 \); \( x - 5=0\) gives \( x = 5 \). So \( B = 1 \) and \( C = 5 \) (or vice versa).

Step3: Find axis of symmetry (D)

For a quadratic function in the form \( f(x)=ax^{2}+bx + c \), the axis of symmetry is given by the formula \( x=-\frac{b}{2a} \).
For the function \( f(x)=-2x^{2}+12x - 10 \), we have \( a=-2 \) and \( b = 12 \).
Substitute into the formula: \( x=-\frac{12}{2\times(-2)}=-\frac{12}{-4}=3 \). So \( D = 3 \).

Answer:

\( A=-10 \), \( B = 1 \), \( C = 5 \), \( D = 3 \)