Quadratic Equation

[Kudash]

Plot the graph of y = -2x²+4x+7, for values of x from x= -2 to x=4. Determine, from the graph, the coordinates and nature of the turning point of the curve. Use the graph to find the roots of

i) -2x²+4x+7=0
ii) 2x² = 4x+14
iii) 4x = 2x²-12
iv) -2x²+x = -6

 

[Deleted member 4993]

Plot the graph of y = -2x²+4x+7, for values of x from x= -2 to x=4. Determine, from the graph, the coordinates and nature of the turning point of the curve. Use the graph to find the roots of

i) -2x²+4x+7=0
ii) 2x² = 4x+14
iii) 4x = 2x²-12
iv) -2x²+x = -6

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

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Please share your work/thoughts about this problem.

 

[Kudash]

I have plotted the graph but I'm very unfamiliar with using the graph to find the roots of an equation when either a, b or c changes

 

[Deleted member 4993]

Determine, from the graph, the coordinates and nature of the turning point

Did you find the turning point/s of y = -2x²+4x+7 ?

What are those?

Now translate or rotate your original graph to depict the other functions.

You can post photo of your or pdf scan of your work.

 

[Dr.Peterson]

I have plotted the graph but I'm very unfamiliar with using the graph to find the roots of an equation when either a, b or c changes

It's not really about changing the coefficients.

Rewrite each equation so that it involves the function [imath]f(x) = -2x^2+4x+7[/imath]; for example the first equation, [imath]-2x^2+4x+7=0[/imath], is [imath]f(x)=0[/imath]. Then look at your graph to see where this equation is true.

I'm not sure, however, how you are expected to answer some of the questions just from the graph, because you can't read off exact values from it in every case. Are approximate values expected?

 

[Steven G]

Note that 2x² = 4x+14 is the same as -2x² + 4x + 14 = 0 is the same as -2x² + 4x + 7 = -7
Given the graph of y = -2x² + 4x + 14, how do you find the solution to -2x² + 4x + 7 = -7

 

[Kudash]

It's not really about changing the coefficients.

Rewrite each equation so that it involves the function [imath]f(x) = -2x^2+4x+7[/imath]; for example the first equation, [imath]-2x^2+4x+7=0[/imath], is [imath]f(x)=0[/imath]. Then look at your graph to see where this equation is true.

I'm not sure, however, how you are expected to answer some of the questions just from the graph, because you can't read off exact values from it in every case. Are approximate values expected?

Good day, yes approximate values are expected

 

[Kudash]

Note that 2x² = 4x+14 is the same as -2x² + 4x + 14 = 0 is the same as -2x² + 4x + 7 = -7
Given the graph of y = -2x² + 4x + 14, how do you find the solution to -2x² + 4x + 7 = -7

Good day, that's where my confusion starts. I'm really not familiar with those