Question

Consider the following quadratic function. Y=x^2-7x+12Find the real zeros, if any, of this function. Reduce all fractions to lowest terms.

Answer 1

`\mleft(3,0\mright)and(4,0)`

Step-by-step explanation

The zeros of a polynomial are the values of x which satisfy the equation y = f(x). Here f(x) is a function of x, and the zeros of the polynomial is the values of x for which the y value is equal to zero

so

Step 1

when the graph crosses the x-axis

let y=0

so

`\begin{gathered} y=x^2-7x+12 \\ \end{gathered}`

The middle number is -7 and the last number is 12.

Factoring means we want something like

`\begin{gathered} \mleft(x+_{}\mright)\mleft(x+_{}\mright) \\ \end{gathered}`

We need two numbers that...

Add together to get -7

Multiply together to get 12

so, the number are

`\begin{gathered} -3\text{ and -4} \\ so \end{gathered}`

`\begin{gathered} y=(x-3_{})(x-4_{}) \\ 0=(x-3_{})(x-4_{}) \end{gathered}`

so, the solutions are

x=3

and

x=4

(3,0) and (4,0)

I hope this helps you