Intersections 62784

A quadratic function is given: y = -x² + 2x + 3
a) determine the intersections with the x, y-axis
and peak V

b) draw a graph and describe

c) for which x applies
f (x) = 3

Correct answer:

Y₀ = [2,0]
Y₁ = [1,0]
X₀ = [0,3]
V = [1,4]
x₁ =  2
x₂ =  0

Step-by-step explanation:

$$\begin{gathered} f\cap y=0 \\ f(x)=0 \\ -X^2+2X+3=3 \\ -X^2+2X=0 \\ {X}^2-2X=0 \\ a=1;b=-2;c=0 \\ D=b^2-4ac=2^2-4\cdot 1\cdot 0=4 \\ D>0 \\ {X}_{1,2}=\frac{-b\pm \sqrt{D}}{2a}=\frac{2\pm \sqrt{4}}{2} \\ {X}_{1,2}=\frac{2\pm 2}{2} \\ {X}_{1,2}=1\pm 1 \\ {X}_1=2 \\ {X}_2=0 \\ {Y}_0=[2,0] \end{gathered}$$

Our quadratic equation calculator calculates it.

$Y_1=[1,0]$

$$\begin{gathered} f\cap x=0 \\ {y}_0=-0^2+2\cdot 0+3=3 \\ {X}_0=[0,3] \end{gathered}$$

$$\begin{gathered} a=-1 \\ b=2 \\ c=3 \\ {V}_x=\frac{-b}{2\cdot a}=\frac{-2}{2\cdot (-1)}=1 \\ {V}_y=c-\frac{b^2}{4\cdot a}=3-\frac{2^2}{4\cdot (-1)}=4 \\ V=[1,4] \end{gathered}$$

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

Our quadratic equation calculator calculates it.

Try another example