I have never come across a question like this before so I have no idea where to start.
Explain why the graph of the equation y = x²+ 4x + 9 does not intersect the x-axis.

Use the discriminant Δ = b2– 4ac
Original post by Blackrose06
I have never come across a question like this before so I have no idea where to start.
Explain why the graph of the equation y = x²+ 4x + 9 does not intersect the x-axis.

Use the discriminant Δ = b2– 4ac

Original post by Blackrose06
I have never come across a question like this before so I have no idea where to start.
Explain why the graph of the equation y = x²+ 4x + 9 does not intersect the x-axis.

Use the discriminant Δ = b2– 4ac

Start by calculating the discriminant using the given formula. b = 4, a = 1 and c = 9.

You should know of the following rules:

If b^2 - 4ac > 0, then the quadratic has 2 real roots

If b^2 - 4ac = 0, then the quadratic has 1 real root

If b^2 - 4ac < 0, then the quadratic has 0 real roots
If Δ is negative, what is happening?

Spoiler

So because the Discriminant is negative, the graph would have no roots, therefore it doesn't touch the X-Axis
Original post by Blackrose06
So because the Discriminant is negative, the graph would have no roots, therefore it doesn't touch the X-Axis

Correct. Though to actually get the mark for writing that statement, I’m fairly certain you would have to show you have actually calculated the discriminant first.
(4)^2 - 4(1)(9) = -20
(edited 6 months ago)
Original post by Blackrose06
So because the Discriminant is negative, the graph would have no roots, therefore it doesn't touch the X-Axis

To be on the safe side, I would say "because the discriminant is negative, the quadratic has no real roots, therefore..."
Ok, thanks guys this has really helped me out (GCSE-> A-level transition work is confusing)
Original post by Blackrose06
Ok, thanks guys this has really helped me out (GCSE-> A-level transition work is confusing)

Is this holiday work before you return for A levels? This sort of stuff is good to reinforce your GCSE algebra and check that you have a good grasp on the concepts before things get more difficult.
Original post by davros
Is this holiday work before you return for A levels? This sort of stuff is good to reinforce your GCSE algebra and check that you have a good grasp on the concepts before things get more difficult.

Yeah. This didn't come up in my GCSE lessons though.
Original post by Blackrose06
I have never come across a question like this before so I have no idea where to start.
Explain why the graph of the equation y = x²+ 4x + 9 does not intersect the x-axis.

Use the discriminant Δ = b2– 4ac

To determine whether the graph of the equation y = + 4x + 9 intersects the x-axis, we need to find the x-values (roots) at which the function y equals zero. In other words, we need to solve for x when y = 0. The graph intersects the x-axis at the points where y is equal to zero.

To find the x-values when y = 0, we set the equation equal to zero:

0 = + 4x + 9

Now, let's try to find the roots of this quadratic equation. We can use the quadratic formula, which is given by:

x = (-b ± √(b² - 4ac)) / 2a

Where the quadratic equation is in the form of ax² + bx + c = 0. In this case, a = 1, b = 4, and c = 9:

x = (-(4) ± √((4)² - 4(1)(9))) / 2(1)
x = (-4 ± √(16 - 36)) / 2
x = (-4 ± √(-20)) / 2

The expression inside the square root, √(-20), is negative. This means the quadratic equation does not have any real roots, and therefore, it does not intersect the x-axis.

The graph of y = + 4x + 9 is a parabola that opens upwards because the coefficient of (1) is positive. As a result, it sits entirely above the x-axis and never crosses it, confirming that it does not have any real x-intercepts or roots. (If you need more guidance or have similar problems you can pm me).
Original post by Blackrose06
I have never come across a question like this before so I have no idea where to start.
Explain why the graph of the equation y = x²+ 4x + 9 does not intersect the x-axis.

Use the discriminant Δ = b2– 4ac

The discriminant is less than 0, b=4 ,a =1 and c =9. The discriminant is (4^2) - 4(1)(9) = 16-36 = -20 since the square root of -20 is 2isqrt5, there are no real solutions and thus, the function doesn't intersect the x-axis.
Original post by π/2=Σk!/(2k+1)!!
The discriminant is less than 0, b=4 ,a =1 and c =9. The discriminant is (4^2) - 4(1)(9) = 16-36 = -20 since the square root of -20 is 2isqrt5, there are no real solutions and thus, the function doesn't intersect the x-axis.

This was answered a week ago, no need to bump it😐.
Original post by π/2=Σk!/(2k+1)!!
The discriminant is less than 0, ...

OK you've done this about 3 times now - could you please take a look at the posting guidelines for this forum https://www.thestudentroom.co.uk/showthread.php?t=4919248 and note particularly that we don't provide full solutions, only suggestions / hints to get the OP started.

In this case, as noted above, the original problem has already been solved, so there was no need to resurrect