Given the function
![f(x)=2x^2+2x-24](https://img.qammunity.org/2023/formulas/mathematics/college/wobwxa83wo2tg8tng4iezfddbujr2h7wbl.png)
To find the x and y intersects of the parabola you have to do as follows:
y-intercept
This is the value of f(x) when x=0, to find it replace the value in the formula:
![\begin{gathered} f(0)=2(0)^2+2\cdot0-24 \\ f(0)=-24 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/l5bnm96p96xvsv3dde9se6ciga6f1dneoq.png)
The y-intercept of the parabola is (0,-24)
Vertex
Calculate the x coordinate using the following formula:
For
![y=ax^2+bx+c](https://img.qammunity.org/2023/formulas/mathematics/high-school/g7mvpjunjwe6qob7ddy7l4f0glbtdi9gci.png)
![x_v=-(b)/(2a)](https://img.qammunity.org/2023/formulas/mathematics/college/kcg8jhoehgrq4behgg4ohzwd4mmzcllet7.png)
For this function:
![x_v=-(2)/(2\cdot2)=-(2)/(4)=-(1)/(2)](https://img.qammunity.org/2023/formulas/mathematics/college/fxdz53cb9qpx3uweok703did5tcazqx3ev.png)
Using this value of x imput it in the formula to reach the value of the y-coordinate of the vertex:
![\begin{gathered} f(x_v=-(1)/(2))=2(-(1)/(2))^2+2(-(1)/(2))-24 \\ f(x_v)=-(49)/(2) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/zqyxtmj6tz9piheyc41art5mbd491h0p7c.png)
The vertex is (-1/2,-49/2)
Using these two points you can draw the function:
Using the graph you can determine the x-intercepts of the function, these are (-4,0) and (3,0)