quadratic equations: y=x^2-x+5 & y=2x+5; solve it in a graph

[Camerontocher]

Hi, i've got an open book assesment and i wrote the question down but im struggling to get to terms of how to do it.
y=x^2-x+5
y=2x+5

i have to solve it in a graph

i know that the answer will be where the two points meet am i correct?

x	-4	-3	-2	-1	0	1	2	3	4
y=x^2-x+5
y=2x+5

i was taught to fill in the gaps using this technique then fill it in a draw the graph on paper.

but in my notes that im allowed to look at in class doesnt tell me what to type in on the calculator to plot the points. the tutor didnt tell us that and to be completely honest with you i dont think ive learnt that much, and since its an assessment he cant help me.

so if anyone could advise me on what to do it would be great thanks.
i wasnt so sure whether the equations above were in the correct rearrangement????

 

[Ishuda]

Hi, i've got an open book assesment and i wrote the question down but im struggling to get to terms of how to do it.
y=x^2-x+5
y=2x+5

i have to solve it in a graph

i know that the answer will be where the two points meet am i correct?

x	-4	-3	-2	-1	0	1	2	3	4
y=x^2-x+5
y=2x+5

i was taught to fill in the gaps using this technique then fill it in a draw the graph on paper.

but in my notes that im allowed to look at in class doesnt tell me what to type in on the calculator to plot the points. the tutor didnt tell us that and to be completely honest with you i dont think ive learnt that much, and since its an assessment he cant help me.

so if anyone could advise me on what to do it would be great thanks.
i wasnt so sure whether the equations above were in the correct rearrangement????

If you want to use a graphing calculator, you must learn how to use it. Look at your instruction manual. BTW, why not just use paper and pencil?

 

[stapel]

y = x^2 - x + 5
y = 2x + 5

have to solve [the non-linear system of equations] in a graph.

[I think] that the [solution] will be [the point] where the two [graphs] meet; am correct?

Yes; the "solution" to a system of equations is the point (or, possibly, points, in non-linear cases) where the graphs of the equations meet or cross. Intersections are solutions.

x	-4	-3	-2	-1	0	1	2	3	4
y = x^2 - x + 5
y = 2x + 5

was taught to fill in the gaps using this technique then fill it in a draw the graph on paper.

Yes; pick values for x (from the above listing, or other values, depending upon what you find useful), plug them into the equations, and simplify to find the corresponding values of y.

...my notes...[don't] tell me what to type in on the calculator to plot the points.

Why rely on a calculator? Do the work yourself, on scratch paper if necessary. The computations aren't difficult. Just because one of the graphs will be a parabola, this doesn't change the methods at all. Just like you picked x-values, plugged them into the equations, and solved for y when you did straight lines; so also you can do for curvy lines like the first equation above.

(And, from what you've learned about graphing in the past, you can easily see one intersection point, just from the line equations. Hint: y-intercepts!)