Intersections 25141

The quadratic function has the formula y = x²-2x-3. Sketch a graph of this function. Find the intersections with the axes. Find the vertex coordinates.

Correct answer:

y₀ =  -3
x₁ =  3
x₂ =  -1
x₃ =  1
y₃ =  -4

Step-by-step explanation:

$$\begin{gathered} f(x)=y=x^2-2x-3 \\ {x}_0=0 \\ {y}_0=x_0^2-2\cdot x_0-3=-3 \\ {P}_y=(0;y_0)=(0;-3) \end{gathered}$$

$$\begin{gathered} y=0 \\ 0=x^2-2\cdot x-3 \\ 0=x^2-2\cdot x-3 \\ -x^2+2x+3=0 \\ {x}^2-2x-3=0 \\ a=1;b=-2;c=-3 \\ D=b^2-4ac=2^2-4\cdot 1\cdot (-3)=16 \\ D>0 \\ {x}_{1,2}=\frac{-b\pm \sqrt{D}}{2a}=\frac{2\pm \sqrt{16}}{2} \\ {x}_{1,2}=\frac{2\pm 4}{2} \\ {x}_{1,2}=1\pm 2 \\ {x}_1=3 \\ {x}_2=-1 \\ {P}_x=(x_1;0)=(3;0) \end{gathered}$$

Our quadratic equation calculator calculates it.

$$\begin{gathered} x_2=(-1)=-1 \\ {P}_x=(x_2;0)=(-1;0) \end{gathered}$$

$$\begin{gathered} a=1 \\ b=-2 \\ c=-3 \\ {x}_3=-b/(2\cdot a)=-(-2)/(2\cdot 1)=1 \end{gathered}$$

$$\begin{gathered} y_3=-b^2/(4\cdot a)+c=-(-2{)}^2/(4\cdot 1)+(-3)=-4 \\ V(x_3;y_3)=(1;-4) \end{gathered}$$

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